W15. Vectors, Matrices, Lines, Planes, Conic Sections, Linear Transformations, Polar Coordinates, Quadric Surfaces

1. Summary

1.1 Vectors and Scalar Product

A vector is a mathematical object characterized by both magnitude and direction. In three-dimensional space, vectors are typically represented as ordered triples \(\vec{v} = (v_1, v_2, v_3)\) or in column form.

1.1.1 Scalar Product (Dot Product)

The scalar product (also called dot product) of two vectors is a fundamental operation that produces a scalar (real number). For vectors \(\vec{a} = (a_1, a_2, a_3)\) and \(\vec{b} = (b_1, b_2, b_3)\), the scalar product is defined algebraically as:

\[\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3\]

Geometrically, the scalar product relates to the angle \(\theta\) between the vectors:

\[\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta\]

where \(|\vec{a}|\) denotes the magnitude (or norm) of vector \(\vec{a}\), calculated as \(|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}\).

1.1.2 Perpendicularity and Orthogonality

Two vectors are perpendicular (or orthogonal) if and only if their scalar product equals zero: \(\vec{a} \cdot \vec{b} = 0\). This property is crucial for testing perpendicularity in geometric problems.

1.1.3 Parallelogram Law

The parallelogram law is an important identity involving scalar products:

\[(\vec{a} + \vec{b})^2 + (\vec{a} - \vec{b})^2 = 2(|\vec{a}|^2 + |\vec{b}|^2)\]

This identity has a geometric interpretation: in a parallelogram, the sum of the squares of the diagonals equals the sum of the squares of all four sides.

1.2 Vector Product (Cross Product)

The vector product (or cross product) of two vectors in three-dimensional space produces a new vector that is perpendicular to both original vectors.

1.2.1 Definition and Calculation

For vectors \(\vec{a} = (a_1, a_2, a_3)\) and \(\vec{b} = (b_1, b_2, b_3)\), the cross product is calculated using a determinant:

\[\vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}\]

This expands to:

\[\vec{a} \times \vec{b} = (a_2b_3 - a_3b_2)\mathbf{i} - (a_1b_3 - a_3b_1)\mathbf{j} + (a_1b_2 - a_2b_1)\mathbf{k}\]

1.2.2 Geometric Meaning

The magnitude of the cross product \(|\vec{a} \times \vec{b}|\) equals the area of the parallelogram formed by vectors \(\vec{a}\) and \(\vec{b}\). Geometrically:

\[|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta\]

where \(\theta\) is the angle between the vectors.

1.2.3 Important Identity

The relationship between dot product and cross product is given by:

\[|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2\]

This identity combines the sine and cosine components of the angle between vectors, utilizing the fundamental trigonometric identity \(\sin^2\theta + \cos^2\theta = 1\).

1.3 Scalar Triple Product

The scalar triple product combines both the cross product and dot product operations.

1.3.1 Definition

For three vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), the scalar triple product is:

\[\vec{a} \cdot (\vec{b} \times \vec{c})\]

This can be computed as a determinant:

\[\vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}\]

1.3.2 Geometric Interpretation

The absolute value of the scalar triple product \(|\vec{a} \cdot (\vec{b} \times \vec{c})|\) represents the volume of the parallelepiped formed by the three vectors.

1.3.3 Coplanarity Test

If the scalar triple product equals zero, then the three vectors are coplanar (lie in the same plane). This provides a test for determining whether four points in space lie on the same plane: if points \(A\), \(B\), \(C\), and \(D\) are given, compute vectors \(\vec{AB}\), \(\vec{AC}\), and \(\vec{AD}\), then check if \(\vec{AB} \cdot (\vec{AC} \times \vec{AD}) = 0\).

1.4 Lines in a Plane

A line in a two-dimensional plane can be represented in several equivalent forms.

1.4.1 General Form

The general form of a line in the plane is:

\[Ax + By + C = 0\]

where \(A\), \(B\), and \(C\) are constants, and \((A, B)\) represents a vector normal (perpendicular) to the line.

1.4.2 Slope-Intercept Form

When \(B \neq 0\), the line can be written as:

\[y = mx + b\]

where \(m = -A/B\) is the slope and \(b = -C/B\) is the y-intercept.

1.4.3 Relationships Between Lines

Two lines \(A_1x + B_1y + C_1 = 0\) and \(A_2x + B_2y + C_2 = 0\) can be:

• Parallel: if \(\frac{A_1}{A_2} = \frac{B_1}{B_2} \neq \frac{C_1}{C_2}\)
• Coincident: if \(\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}\)
• Intersecting: if \(\frac{A_1}{A_2} \neq \frac{B_1}{B_2}\)
• Perpendicular: if \(A_1A_2 + B_1B_2 = 0\)

1.4.4 Distance from Point to Line

The distance from a point \(P(x_0, y_0)\) to a line \(Ax + By + C = 0\) is given by:

\[d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}\]

1.4.5 Projection of a Point

To find the projection of a point onto a line, construct a perpendicular line through the point to the given line, then find their intersection.

1.5 Planes in Space

A plane in three-dimensional space is a flat, two-dimensional surface extending infinitely.

1.5.1 General Form

The general equation of a plane is:

\[Ax + By + Cz + D = 0\]

where \((A, B, C)\) is a normal vector to the plane (perpendicular to every vector lying in the plane).

1.5.2 Point-Normal Form

If a plane passes through point \(M_0(x_0, y_0, z_0)\) with normal vector \(\vec{n} = (A, B, C)\), its equation is:

\[A(x - x_0) + B(y - y_0) + C(z - z_0) = 0\]

1.5.3 Three-Point Form

To find the equation of a plane through three points \(M_1\), \(M_2\), and \(M_3\):

1. Compute vectors \(\vec{M_1M_2}\) and \(\vec{M_1M_3}\)
2. Find the normal vector: \(\vec{n} = \vec{M_1M_2} \times \vec{M_1M_3}\)
3. Use the point-normal form with any of the three points

1.5.4 Mutual Arrangement of Planes

The relationship between three planes can be determined by examining the determinant of their normal vectors. If the determinant is non-zero, the three planes intersect at a single point.

1.5.5 Distance from Point to Plane

The distance from point \(P(x_0, y_0, z_0)\) to plane \(Ax + By + Cz + D = 0\) is:

\[d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}\]

1.6 Lines in Space

Lines in three-dimensional space require more complex representations than lines in a plane.

1.6.1 Canonical Form

The canonical (or symmetric) form of a line through point \(M_0(x_0, y_0, z_0)\) with direction vector \(\vec{v} = (l, m, n)\) is:

\[\frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n}\]

1.6.2 Parametric Form

The parametric equations of the same line are:

\[\begin{cases} x = x_0 + lt \\ y = y_0 + mt \\ z = z_0 + nt \end{cases}\]

where \(t \in \mathbb{R}\) is the parameter.

1.6.3 Line as Intersection of Planes

A line can also be represented as the intersection of two planes:

\[\begin{cases} A_1x + B_1y + C_1z + D_1 = 0 \\ A_2x + B_2y + C_2z + D_2 = 0 \end{cases}\]

To convert to canonical form, find the direction vector as \(\vec{v} = \vec{n}_1 \times \vec{n}_2\) and find any point on the line by setting one coordinate to a convenient value.

1.6.4 Angle Between Lines

The angle \(\theta\) between two lines with direction vectors \(\vec{v}_1\) and \(\vec{v}_2\) is:

\[\cos \theta = \frac{|\vec{v}_1 \cdot \vec{v}_2|}{|\vec{v}_1| |\vec{v}_2|}\]

Note that we take the absolute value to get the acute angle.

1.7 Conic Sections (Second-Order Curves)

Conic sections are curves obtained by intersecting a plane with a double cone. They include ellipses, parabolas, and hyperbolas.

1.7.1 General Equation

The general equation of a conic section is:

\[Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\]

1.7.2 Classification by Discriminant

The discriminant \(\Delta = B^2 - 4AC\) determines the type of conic:

• \(\Delta < 0\): Ellipse (or circle if \(A = C\) and \(B = 0\))
• \(\Delta = 0\): Parabola
• \(\Delta > 0\): Hyperbola

1.7.3 Ellipse

An ellipse is the set of all points where the sum of distances to two fixed points (called foci) is constant.

Standard form: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) where \(a > b > 0\)

• Semi-major axis: \(a\) (half the length of the longer axis)
• Semi-minor axis: \(b\) (half the length of the shorter axis)
• Eccentricity: \(e = \frac{c}{a}\) where \(c^2 = a^2 - b^2\) and \(0 < e < 1\)
• Foci: Located at \((\pm c, 0)\)
• Directrices: Lines \(x = \pm \frac{a}{e}\)

1.7.4 Hyperbola

A hyperbola is the set of all points where the absolute difference of distances to two foci is constant.

Standard form: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)

• Semi-transverse axis: \(a\)
• Semi-conjugate axis: \(b\)
• Eccentricity: \(e = \frac{c}{a}\) where \(c^2 = a^2 + b^2\) and \(e > 1\)
• Foci: Located at \((\pm c, 0)\)
• Directrices: Lines \(x = \pm \frac{a}{e}\)
• Asymptotes: Lines \(y = \pm \frac{b}{a}x\)

1.7.5 Parabola

A parabola is the set of all points equidistant from a fixed point (the focus) and a fixed line (the directrix).

Standard form: \(y^2 = 2px\) where \(p > 0\)

• Focus: Located at \((p/2, 0)\)
• Directrix: Line \(x = -p/2\)
• Eccentricity: \(e = 1\) (always for parabolas)
• Vertex: The point closest to the directrix, at the origin \((0, 0)\) in standard form

1.7.6 Rotation of Axes

When a conic has an \(xy\) term (\(B \neq 0\)), we can eliminate it by rotating the coordinate axes. The rotation angle \(\theta\) is found from:

\[\cot 2\theta = \frac{A - C}{B}\]

The rotation transformation is:

\[\begin{cases} x = x'\cos\theta - y'\sin\theta \\ y = x'\sin\theta + y'\cos\theta \end{cases}\]

For \(\theta = 45°\), this simplifies to:

\[\begin{cases} x = \frac{1}{\sqrt{2}}(x' - y') \\ y = \frac{1}{\sqrt{2}}(x' + y') \end{cases}\]

1.8 Polar Coordinates

Polar coordinates provide an alternative way to specify points in a plane using distance and angle rather than Cartesian coordinates.

1.8.1 Definition

A point \(P\) is specified by:

• \(r\) (or \(\rho\)): the distance from the origin to \(P\)
• \(\theta\) (or \(\varphi\)): the angle from the positive x-axis to the line segment connecting the origin to \(P\)

1.8.2 Conversion Formulas

To convert between Cartesian \((x, y)\) and polar \((r, \theta)\) coordinates:

• From polar to Cartesian: \(x = r\cos\theta\), \(y = r\sin\theta\)
• From Cartesian to polar: \(r = \sqrt{x^2 + y^2}\), \(\theta = \arctan(y/x)\)

1.8.3 Conics in Polar Coordinates

Conics with a focus at the origin have the form:

\[r = \frac{ed}{1 + e\cos(\theta - \theta_0)}\]

where \(e\) is the eccentricity and \(d\) is a parameter related to the directrix.

1.9 Matrices

A matrix is a rectangular array of numbers arranged in rows and columns.

1.9.1 Basic Definitions

An \(m \times n\) matrix has \(m\) rows and \(n\) columns:

\[A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix}\]

Special matrices:

• Square matrix: \(m = n\)
• Identity matrix \(I\): Square matrix with 1s on the diagonal and 0s elsewhere
• Zero matrix: All entries are 0

1.9.2 Matrix Operations

Addition: Add corresponding entries (matrices must have same dimensions)

Scalar multiplication: Multiply every entry by the scalar

Matrix multiplication: For \(A\) (\(m \times p\)) and \(B\) (\(p \times n\)), the product \(AB\) is \(m \times n\) with entries:

\[(AB)_{ij} = \sum_{k=1}^{p} a_{ik}b_{kj}\]

Note: Matrix multiplication is not commutative: generally \(AB \neq BA\).

Transpose: \(A^T\) is formed by swapping rows and columns: \((A^T)_{ij} = A_{ji}\)

1.9.3 Determinant

The determinant is a scalar value associated with square matrices. For a \(2 \times 2\) matrix:

\[\det\begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc\]

For larger matrices, use cofactor expansion along a row or column.

1.9.4 Matrix Inverse

A square matrix \(A\) has an inverse \(A^{-1}\) if \(AA^{-1} = A^{-1}A = I\). The inverse exists if and only if \(\det(A) \neq 0\).

For a \(2 \times 2\) matrix:

\[\begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\]

1.9.5 Matrix Rank

The rank of a matrix is the maximum number of linearly independent rows (or columns). It can be found using Gaussian elimination.

1.10 Linear Independence and Basis

1.10.1 Linear Independence

Vectors \(\vec{v}_1, \vec{v}_2, \ldots, \vec{v}_n\) are linearly independent if the only solution to:

\[c_1\vec{v}_1 + c_2\vec{v}_2 + \cdots + c_n\vec{v}_n = \vec{0}\]

is \(c_1 = c_2 = \cdots = c_n = 0\). Otherwise, they are linearly dependent.

1.10.2 Basis

A basis for a vector space is a set of linearly independent vectors that span the space. The standard basis for \(\mathbb{R}^3\) is \(\{\hat{i}, \hat{j}, \hat{k}\} = \{(1,0,0), (0,1,0), (0,0,1)\}\).

1.10.3 Change of Basis

To convert coordinates from one basis to another, use a transition matrix \(P\). If \([\vec{x}]_B\) represents vector \(\vec{x}\) in basis \(B\) and \([\vec{x}]_E\) in standard basis \(E\):

\[[\vec{x}]_E = P_{E \leftarrow B}[\vec{x}]_B\]

The transition matrix \(P_{E \leftarrow B}\) has the basis vectors of \(B\) as its columns (expressed in basis \(E\)).

1.11 Subspaces

A subspace of \(\mathbb{R}^n\) is a subset that is closed under addition and scalar multiplication.

1.11.1 Subspace Criteria

For a set \(W\) to be a subspace:

1. The zero vector must be in \(W\)
2. If \(\vec{u}, \vec{v} \in W\), then \(\vec{u} + \vec{v} \in W\) (closure under addition)
3. If \(\vec{v} \in W\) and \(c\) is a scalar, then \(c\vec{v} \in W\) (closure under scalar multiplication)

1.11.2 Common Examples

• Planes through the origin in \(\mathbb{R}^3\) are subspaces
• Lines through the origin are subspaces
• Planes or lines not through the origin are NOT subspaces

1.12 Linear Transformations

A linear transformation \(T: \mathbb{R}^n \to \mathbb{R}^m\) is a function that preserves vector addition and scalar multiplication.

1.12.1 Definition

\(T\) is linear if:

1. \(T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})\) for all vectors \(\vec{u}, \vec{v}\)
2. \(T(c\vec{v}) = cT(\vec{v})\) for all vectors \(\vec{v}\) and scalars \(c\)

1.12.2 Matrix Representation

Every linear transformation can be represented by a matrix. If \(T: \mathbb{R}^n \to \mathbb{R}^m\), there exists an \(m \times n\) matrix \([T]\) such that:

\[T(\vec{x}) = [T]\vec{x}\]

1.12.3 Kernel and Image

For a linear transformation \(T\):

• The kernel (or null space) \(\text{ker}(T)\) is the set of all vectors that map to zero: \(\text{ker}(T) = \{\vec{x} : T(\vec{x}) = \vec{0}\}\)
• The image (or range) \(\text{im}(T)\) is the set of all possible outputs: \(\text{im}(T) = \{T(\vec{x}) : \vec{x} \in \mathbb{R}^n\}\)

1.12.4 Rank-Nullity Theorem

For a linear transformation \(T: \mathbb{R}^n \to \mathbb{R}^m\):

\[\dim(\text{ker}(T)) + \dim(\text{im}(T)) = n\]

where \(\dim(\text{im}(T))\) is the rank and \(\dim(\text{ker}(T))\) is the nullity.

1.13 Quadric Surfaces

Quadric surfaces are three-dimensional surfaces defined by second-degree equations in \(x\), \(y\), and \(z\).

1.13.1 Ellipsoid

\[\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\]

An ellipsoid is a closed, bounded surface resembling a stretched sphere. When \(a = b = c\), it’s a sphere.

1.13.2 Hyperboloid of One Sheet

\[\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1\]

This surface is connected and extends infinitely along the \(z\)-axis, with a “waist” at the center.

1.13.3 Hyperboloid of Two Sheets

\[\frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1\]

This consists of two separate sheets, one for \(x \geq a\) and one for \(x \leq -a\).

1.13.4 Elliptic Paraboloid

\[z = \frac{x^2}{a^2} + \frac{y^2}{b^2}\]

This surface opens upward (or downward if negative) and has elliptical cross-sections parallel to the \(xy\)-plane.

1.13.5 Hyperbolic Paraboloid

\[z = \frac{x^2}{a^2} - \frac{y^2}{b^2}\]

This is a “saddle-shaped” surface, curving upward in one direction and downward in the perpendicular direction.

1.13.6 Cone

\[\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 0\]

A cone with vertex at the origin. The right side is 0, distinguishing it from hyperboloids.

1.13.7 Cross-Sections

To analyze quadric surfaces, examine their intersection with coordinate planes or planes parallel to them. For example, setting \(x = k\) (constant) in an ellipsoid equation gives an ellipse in the plane \(x = k\).

1.14 Systems of Linear Equations

1.14.1 Gaussian Elimination

Gaussian elimination is a method to solve systems of linear equations by converting the augmented matrix to row echelon form using elementary row operations:

1. Swap two rows
2. Multiply a row by a non-zero scalar
3. Add a multiple of one row to another row

1.14.2 Cramer’s Rule

For a system \(A\vec{x} = \vec{b}\) where \(A\) is \(n \times n\) and \(\det(A) \neq 0\), the solution is:

\[x_i = \frac{\det(A_i)}{\det(A)}\]

where \(A_i\) is matrix \(A\) with column \(i\) replaced by vector \(\vec{b}\).

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2. Definitions

• Vector: A mathematical object with both magnitude and direction, typically represented as an ordered tuple of numbers.
• Scalar Product (Dot Product): An operation that takes two vectors and returns a scalar: \(\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3\).
• Magnitude (Norm): The length of a vector: \(|\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}\).
• Orthogonal (Perpendicular): Two vectors are orthogonal if their dot product is zero.
• Vector Product (Cross Product): An operation on two vectors in 3D that produces a third vector perpendicular to both: \(\vec{a} \times \vec{b}\).
• Scalar Triple Product: The operation \(\vec{a} \cdot (\vec{b} \times \vec{c})\), which gives the volume of a parallelepiped.
• Coplanar: Vectors or points that lie in the same plane.
• Line: A one-dimensional geometric object extending infinitely in both directions.
• Plane: A flat, two-dimensional surface in three-dimensional space.
• Normal Vector: A vector perpendicular to a line (in 2D) or plane (in 3D).
• Direction Vector: A vector indicating the direction of a line.
• Parallel Lines: Lines that never intersect (same direction vector or proportional normal vectors).
• Coincident Lines: Lines that overlap completely (identical equations).
• Skew Lines: Non-parallel lines in 3D space that do not intersect.
• Conic Section: A curve obtained by intersecting a plane with a cone (ellipse, parabola, or hyperbola).
• Ellipse: The locus of points where the sum of distances to two foci is constant.
• Hyperbola: The locus of points where the absolute difference of distances to two foci is constant.
• Parabola: The locus of points equidistant from a focus and a directrix.
• Eccentricity: A measure of how much a conic section deviates from being circular (\(e < 1\) for ellipse, \(e = 1\) for parabola, \(e > 1\) for hyperbola).
• Focus (Foci): Fixed point(s) used in the definition of conic sections.
• Directrix: A fixed line used in the definition of conic sections.
• Asymptote: A line that a curve approaches but never reaches (relevant for hyperbolas).
• Discriminant: For a conic equation \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\), the discriminant is \(\Delta = B^2 - 4AC\), which determines the conic type.
• Polar Coordinates: A coordinate system where points are specified by distance from origin and angle from reference axis.
• Matrix: A rectangular array of numbers organized in rows and columns.
• Square Matrix: A matrix with equal number of rows and columns.
• Identity Matrix: A square matrix with 1s on the diagonal and 0s elsewhere.
• Determinant: A scalar value associated with square matrices that indicates invertibility.
• Matrix Inverse: For matrix \(A\), the inverse \(A^{-1}\) satisfies \(AA^{-1} = I\).
• Transpose: The matrix \(A^T\) obtained by swapping rows and columns of \(A\).
• Matrix Rank: The maximum number of linearly independent rows or columns in a matrix.
• Linear Independence: Vectors are linearly independent if no vector can be expressed as a linear combination of the others.
• Linear Dependence: Vectors are linearly dependent if at least one can be expressed as a linear combination of the others.
• Basis: A set of linearly independent vectors that span a vector space.
• Standard Basis: The basis consisting of unit vectors along coordinate axes.
• Transition Matrix: A matrix that converts coordinates from one basis to another.
• Subspace: A subset of a vector space that is closed under addition and scalar multiplication.
• Linear Transformation: A function between vector spaces that preserves addition and scalar multiplication.
• Kernel (Null Space): The set of vectors that a linear transformation maps to zero.
• Image (Range): The set of all possible outputs of a linear transformation.
• Rank-Nullity Theorem: States that dimension of kernel plus dimension of image equals dimension of domain.
• Quadric Surface: A three-dimensional surface defined by a second-degree equation in three variables.
• Ellipsoid: A closed quadric surface; all three squared terms have the same sign.
• Hyperboloid: A quadric surface with one or two sheets; squared terms have mixed signs.
• Paraboloid: A quadric surface where one variable appears to the first degree; unbounded in one direction.
• Cone: A quadric surface with vertex at origin; equation equals zero.
• Gaussian Elimination: A method for solving systems of linear equations using row operations.
• Cramer’s Rule: A formula for solving systems using determinants.
• Augmented Matrix: A matrix formed by appending the constant vector to the coefficient matrix of a linear system.

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3. Formulas

• Scalar Product: \(\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 = |\vec{a}||\vec{b}|\cos\theta\)
• Vector Magnitude: \(|\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}\)
• Perpendicularity Condition: \(\vec{a} \perp \vec{b} \Leftrightarrow \vec{a} \cdot \vec{b} = 0\)
• Parallelogram Law: \((\vec{a} + \vec{b})^2 + (\vec{a} - \vec{b})^2 = 2(|\vec{a}|^2 + |\vec{b}|^2)\)
• Vector Product: \(\vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}\)
• Cross Product Magnitude: \(|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta\) (area of parallelogram)
• Dot and Cross Product Identity: \(|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2|\vec{b}|^2\)
• Scalar Triple Product: \(\vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}\) (volume of parallelepiped)
• Coplanarity Test: Four points \(A, B, C, D\) are coplanar if \(\vec{AB} \cdot (\vec{AC} \times \vec{AD}) = 0\)
• Distance from Point to Line (2D): \(d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}\) for point \((x_0, y_0)\) and line \(Ax + By + C = 0\)
• Perpendicularity of Lines (2D): Lines \(A_1x + B_1y + C_1 = 0\) and \(A_2x + B_2y + C_2 = 0\) are perpendicular if \(A_1A_2 + B_1B_2 = 0\)
• Plane Equation (Point-Normal Form): \(A(x - x_0) + B(y - y_0) + C(z - z_0) = 0\) for plane through \((x_0, y_0, z_0)\) with normal \(\vec{n} = (A, B, C)\)
• Distance from Point to Plane: \(d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}\) for point \((x_0, y_0, z_0)\) and plane \(Ax + By + Cz + D = 0\)
• Line in Space (Canonical Form): \(\frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n}\) for line through \((x_0, y_0, z_0)\) with direction \((l, m, n)\)
• Line in Space (Parametric Form): \(x = x_0 + lt\), \(y = y_0 + mt\), \(z = z_0 + nt\), \(t \in \mathbb{R}\)
• Angle Between Lines: \(\cos\theta = \frac{|\vec{v}_1 \cdot \vec{v}_2|}{|\vec{v}_1||\vec{v}_2|}\) for lines with direction vectors \(\vec{v}_1\) and \(\vec{v}_2\)
• Conic Discriminant: \(\Delta = B^2 - 4AC\) for \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\)
• Ellipse Standard Form: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) with \(a > b > 0\)
• Ellipse Eccentricity: \(e = \frac{c}{a}\) where \(c^2 = a^2 - b^2\), \(0 < e < 1\)
• Hyperbola Standard Form: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
• Hyperbola Eccentricity: \(e = \frac{c}{a}\) where \(c^2 = a^2 + b^2\), \(e > 1\)
• Hyperbola Asymptotes: \(y = \pm\frac{b}{a}x\) for standard form centered at origin
• Parabola Standard Form: \(y^2 = 2px\) or \(x^2 = 2py\)
• Parabola Eccentricity: \(e = 1\) (always)
• Rotation Angle for Conics: \(\cot 2\theta = \frac{A - C}{B}\) to eliminate \(xy\) term
• Rotation Transformation: \(x = x'\cos\theta - y'\sin\theta\), \(y = x'\sin\theta + y'\cos\theta\)
• Polar to Cartesian: \(x = r\cos\theta\), \(y = r\sin\theta\)
• Cartesian to Polar: \(r = \sqrt{x^2 + y^2}\), \(\theta = \arctan(y/x)\)
• Conic in Polar Form: \(r = \frac{ed}{1 + e\cos\theta}\) for conic with focus at origin
• Matrix Multiplication: \((AB)_{ij} = \sum_{k} A_{ik}B_{kj}\)
• Determinant (2×2): \(\det\begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc\)
• Matrix Inverse (2×2): \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\)
• Change of Basis: \([\vec{x}]_E = P_{E \leftarrow B}[\vec{x}]_B\) where \(P\) has basis vectors as columns
• Rank-Nullity Theorem: \(\dim(\ker(T)) + \dim(\text{im}(T)) = \dim(\text{domain})\)
• Cramer’s Rule: \(x_i = \frac{\det(A_i)}{\det(A)}\) where \(A_i\) has column \(i\) replaced by constant vector
• Ellipsoid: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\)
• Hyperboloid of One Sheet: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1\)
• Hyperboloid of Two Sheets: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1\)
• Elliptic Paraboloid: \(z = \frac{x^2}{a^2} + \frac{y^2}{b^2}\)
• Hyperbolic Paraboloid: \(z = \frac{x^2}{a^2} - \frac{y^2}{b^2}\)
• Cone: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 0\)

---

4. Examples

4.1. Perpendicularity of Vectors Using Scalar Product (Final Recap, Task 1)

Prove that vectors \(\vec{d} = \vec{a}(\vec{b} \cdot \vec{c}) - \vec{b}(\vec{a} \cdot \vec{c})\) and \(\vec{c}\) are perpendicular.

Click to see the solution

Key Concept: Two vectors are perpendicular if and only if their scalar product equals zero.

1. Compute the scalar product \(\vec{d} \cdot \vec{c}\):

\[\vec{d} \cdot \vec{c} = [\vec{a}(\vec{b} \cdot \vec{c}) - \vec{b}(\vec{a} \cdot \vec{c})] \cdot \vec{c}\]

2. Distribute the dot product:

\[= \vec{a} \cdot \vec{c} \cdot (\vec{b} \cdot \vec{c}) - \vec{b} \cdot \vec{c} \cdot (\vec{a} \cdot \vec{c})\]

3. Note that scalar products are commutative: Both terms are equal (they’re just products of scalars), so they cancel:

\[= (\vec{b} \cdot \vec{c})(\vec{a} \cdot \vec{c}) - (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{c}) = 0\]

4. Conclusion: Since \(\vec{d} \cdot \vec{c} = 0\), we have \(\vec{d} \perp \vec{c}\).

Answer: The vectors are perpendicular since their scalar product equals zero.

4.2. Parallelogram Law Verification (Final Recap, Task 2)

Verify \((\vec{a} + \vec{b})^2 + (\vec{a} - \vec{b})^2 = 2(|\vec{a}|^2 + |\vec{b}|^2)\) and give a geometric interpretation.

Click to see the solution

Key Concept: The square of a vector \(\vec{v}^2\) means \(\vec{v} \cdot \vec{v} = |\vec{v}|^2\).

1. Expand \((\vec{a} + \vec{b})^2\):

\[(\vec{a} + \vec{b})^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2\]

2. Expand \((\vec{a} - \vec{b})^2\):

\[(\vec{a} - \vec{b})^2 = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) = |\vec{a}|^2 - 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2\]

3. Add the two expansions:

\[(\vec{a} + \vec{b})^2 + (\vec{a} - \vec{b})^2 = |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 + |\vec{a}|^2 - 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2\]

4. Simplify (the \(2(\vec{a} \cdot \vec{b})\) terms cancel):

\[= 2|\vec{a}|^2 + 2|\vec{b}|^2 = 2(|\vec{a}|^2 + |\vec{b}|^2)\]

Geometric Interpretation: In a parallelogram formed by vectors \(\vec{a}\) and \(\vec{b}\), the diagonals are \(\vec{a} + \vec{b}\) and \(\vec{a} - \vec{b}\). The sides have lengths \(|\vec{a}|\) and \(|\vec{b}|\). This identity states that the sum of the squares of the diagonal lengths equals the sum of the squares of all four side lengths.

Answer: Identity verified; it represents the parallelogram law.

4.3. Perpendicular Diagonals (Final Recap, Task 3)

Given \(A(1, -2, 2)\), \(B(1, 4, 0)\), \(C(-4, 1, 1)\), \(D(-5, -5, 3)\). Prove that diagonals \(AC\) and \(BD\) are perpendicular.

Click to see the solution

1. Find vector \(\vec{AC}\):

\[\vec{AC} = C - A = (-4-1, 1-(-2), 1-2) = (-5, 3, -1)\]

2. Find vector \(\vec{BD}\):

\[\vec{BD} = D - B = (-5-1, -5-4, 3-0) = (-6, -9, 3)\]

3. Compute the scalar product:

\[\vec{AC} \cdot \vec{BD} = (-5)(-6) + (3)(-9) + (-1)(3) = 30 - 27 - 3 = 0\]

4. Conclusion: Since \(\vec{AC} \cdot \vec{BD} = 0\), the diagonals are perpendicular.

Answer: The diagonals \(AC\) and \(BD\) are perpendicular.

4.4. Area of Parallelogram (Final Recap, Task 4)

Find the area of the parallelogram formed by vectors \(\vec{a} = (8, 4, 1)\) and \(\vec{b} = (2, -1, 1)\).

Click to see the solution

Key Concept: The area of a parallelogram formed by two vectors equals the magnitude of their cross product.

1. Compute the cross product \(\vec{a} \times \vec{b}\):

\[\vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 8 & 4 & 1 \\ 2 & -1 & 1 \end{vmatrix}\]

2. Expand the determinant:

\[= \mathbf{i}(4 \cdot 1 - 1 \cdot (-1)) - \mathbf{j}(8 \cdot 1 - 1 \cdot 2) + \mathbf{k}(8 \cdot (-1) - 4 \cdot 2)\]

\[= \mathbf{i}(4 + 1) - \mathbf{j}(8 - 2) + \mathbf{k}(-8 - 8)\]

\[= 5\mathbf{i} - 6\mathbf{j} - 16\mathbf{k} = (5, -6, -16)\]

3. Find the magnitude:

\[|\vec{a} \times \vec{b}| = \sqrt{5^2 + (-6)^2 + (-16)^2} = \sqrt{25 + 36 + 256} = \sqrt{317}\]

Answer: The area of the parallelogram is \(\sqrt{317}\) square units.

4.5. Identity Relating Dot and Cross Products (Final Recap, Task 5)

Prove \(|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2\).

Click to see the solution

Key Concept: Use the geometric interpretations of dot and cross products in terms of the angle between vectors.

1. Express cross product magnitude:

\[|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta\]

where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\).

2. Express dot product:

\[\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta\]

3. Square both expressions:

\[|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2\sin^2\theta\]

\[(\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2|\vec{b}|^2\cos^2\theta\]

4. Add them together:

\[|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2|\vec{b}|^2\sin^2\theta + |\vec{a}|^2|\vec{b}|^2\cos^2\theta\]

5. Factor out common terms:

\[= |\vec{a}|^2|\vec{b}|^2(\sin^2\theta + \cos^2\theta)\]

6. Apply the Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\):

\[= |\vec{a}|^2|\vec{b}|^2\]

Answer: The identity is proven using the geometric definitions and the Pythagorean identity.

4.6. Coplanarity of Four Points (Final Recap, Task 6)

Prove that points \(A(1, 2, -1)\), \(B(0, 1, 5)\), \(C(-1, 2, 1)\), \(D(2, 1, 3)\) lie on the same plane.

Click to see the solution

Key Concept: Four points are coplanar if and only if the scalar triple product of three vectors connecting them equals zero.

1. Find vectors from point \(A\) to the other points:

\[\vec{AB} = B - A = (0-1, 1-2, 5-(-1)) = (-1, -1, 6)\]

\[\vec{AC} = C - A = (-1-1, 2-2, 1-(-1)) = (-2, 0, 2)\]

\[\vec{AD} = D - A = (2-1, 1-2, 3-(-1)) = (1, -1, 4)\]

2. Compute \(\vec{AC} \times \vec{AD}\):

\[\vec{AC} \times \vec{AD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 0 & 2 \\ 1 & -1 & 4 \end{vmatrix}\]

\[= \mathbf{i}(0 \cdot 4 - 2 \cdot (-1)) - \mathbf{j}((-2) \cdot 4 - 2 \cdot 1) + \mathbf{k}((-2) \cdot (-1) - 0 \cdot 1)\]

\[= \mathbf{i}(2) - \mathbf{j}(-10) + \mathbf{k}(2) = (2, 10, 2)\]

3. Compute the scalar triple product:

\[\vec{AB} \cdot (\vec{AC} \times \vec{AD}) = (-1, -1, 6) \cdot (2, 10, 2)\]

\[= (-1)(2) + (-1)(10) + (6)(2) = -2 - 10 + 12 = 0\]

4. Conclusion: Since the scalar triple product equals zero, the four points are coplanar.

Answer: The points lie on the same plane.

4.7. Projection of Point onto Line (Final Recap, Task 7)

Find the projection of point \((-5, 6)\) onto the line \(7x - 13y - 105 = 0\).

Click to see the solution

Key Concept: The projection of a point onto a line is found by constructing a perpendicular from the point to the line and finding the intersection.

1. Find the perpendicular direction: The line has normal vector \((7, -13)\), so its direction vector is \((13, 7)\) (perpendicular to the normal). The perpendicular to this line has direction \((-13, 7)\) or equivalently slope \(-13/7\).
2. Write the equation of the perpendicular line through \((-5, 6)\):

Using point-slope form with slope \(-13/7\):

\[y - 6 = -\frac{13}{7}(x - (-5))\]

\[y - 6 = -\frac{13}{7}(x + 5)\]

\[7(y - 6) = -13(x + 5)\]

\[7y - 42 = -13x - 65\]

\[13x + 7y + 23 = 0\]

3. Solve the system of the original line and the perpendicular:

\[\begin{cases} 7x - 13y - 105 = 0 \\ 13x + 7y + 23 = 0 \end{cases}\]

Multiply first equation by 7: \(49x - 91y - 735 = 0\)

Multiply second equation by 13: \(169x + 91y + 299 = 0\)

Add them: \(218x - 436 = 0\), so \(x = 2\)

4. Find \(y\): Substitute \(x = 2\) into \(13x + 7y + 23 = 0\):

\[13(2) + 7y + 23 = 0\]

\[26 + 7y + 23 = 0\]

\[7y = -49\]

\[y = -7\]

Answer: The projection is \((2, -7)\).

4.8. Relationship Between Lines (Part 1) (Final Recap, Task 8a)

Determine the relationship between the lines \(x - 2y + 4 = 0\) and \(-2x + 4y - 8 = 0\).

Click to see the solution

1. Divide the second equation by \(-2\):

\[\frac{-2x + 4y - 8}{-2} = \frac{0}{-2}\]

\[x - 2y + 4 = 0\]

2. Compare: The second equation becomes identical to the first equation.

Answer: The lines are coincident (they are the same line).

4.9. Relationship Between Lines (Part 2) (Final Recap, Task 8b)

Determine the relationship between the lines \(x + y + 5 = 0\) and \(2x + 3y + 10 = 0\).

Click to see the solution

1. Check if parallel: Compare ratios \(\frac{A_1}{A_2} = \frac{1}{2}\) and \(\frac{B_1}{B_2} = \frac{1}{3}\).

Since \(\frac{1}{2} \neq \frac{1}{3}\), the lines are not parallel.

2. Find intersection: Multiply first equation by 2:

\[2x + 2y + 10 = 0\]

Subtract from second equation:

\[2x + 3y + 10 - (2x + 2y + 10) = 0\]

\[y = 0\]

Substitute into first equation: \(x + 0 + 5 = 0\), so \(x = -5\).

Answer: The lines intersect at point \((-5, 0)\).

4.10. Relationship Between Lines (Part 3) (Final Recap, Task 8c)

Determine the relationship between the lines \(4x + 6y - 7 = 0\) and \(2x + 3y - 1 = 0\).

Click to see the solution

1. Multiply the second equation by 2:

\[2(2x + 3y - 1) = 0\]

\[4x + 6y - 2 = 0\]

2. Compare with the first equation \(4x + 6y - 7 = 0\):

The coefficients of \(x\) and \(y\) match (\(\frac{4}{4} = \frac{6}{6} = 1\)), but \(\frac{-7}{-2} \neq 1\).

Answer: The lines are parallel (they have the same direction but different \(y\)-intercepts).

4.11. Perpendicular Lines (Part 1) (Final Recap, Task 9a)

Determine if lines \(3x + 7y + 4 = 0\) and \(7x - 3y + 2 = 0\) are perpendicular.

Click to see the solution

Key Concept: Lines \(A_1x + B_1y + C_1 = 0\) and \(A_2x + B_2y + C_2 = 0\) are perpendicular if \(A_1A_2 + B_1B_2 = 0\).

1. Apply the perpendicularity condition:

\[A_1A_2 + B_1B_2 = (3)(7) + (7)(-3) = 21 - 21 = 0\]

Answer: Yes, the lines are perpendicular.

4.12. Perpendicular Lines (Part 2) (Final Recap, Task 9b)

Determine if lines \(5x + 6y - 8 = 0\) and \(6x + 5y + 2 = 0\) are perpendicular.

Click to see the solution

1. Apply the perpendicularity condition:

\[A_1A_2 + B_1B_2 = (5)(6) + (6)(5) = 30 + 30 = 60 \neq 0\]

Answer: No, the lines are not perpendicular.

4.13. Lines Parallel to Triangle Sides (Final Recap, Task 10)

For triangle with vertices \(A(-1, 2)\), \(B(3, -1)\), \(C(0, 4)\), find the equation of the line through each vertex that is parallel to the opposite side.

Click to see the solution

(a) Line through \(A\) parallel to \(BC\):

1. Find slope of \(BC\):

\[m_{BC} = \frac{4 - (-1)}{0 - 3} = \frac{5}{-3} = -\frac{5}{3}\]

2. Equation through \(A(-1, 2)\) with slope \(-5/3\):

\[y - 2 = -\frac{5}{3}(x - (-1))\]

\[3(y - 2) = -5(x + 1)\]

\[3y - 6 = -5x - 5\]

\[5x + 3y - 1 = 0\]

(b) Line through \(B\) parallel to \(AC\):

1. Find slope of \(AC\):

\[m_{AC} = \frac{4 - 2}{0 - (-1)} = \frac{2}{1} = 2\]

2. Equation through \(B(3, -1)\) with slope \(2\):

\[y - (-1) = 2(x - 3)\]

\[y + 1 = 2x - 6\]

\[y = 2x - 7\]

(c) Line through \(C\) parallel to \(AB\):

1. Find slope of \(AB\):

\[m_{AB} = \frac{-1 - 2}{3 - (-1)} = \frac{-3}{4} = -\frac{3}{4}\]

2. Equation through \(C(0, 4)\) with slope \(-3/4\):

\[y - 4 = -\frac{3}{4}(x - 0)\]

\[4(y - 4) = -3x\]

\[4y - 16 = -3x\]

\[3x + 4y - 16 = 0\]

Answer:

• Line through \(A\): \(5x + 3y - 1 = 0\)
• Line through \(B\): \(y = 2x - 7\)
• Line through \(C\): \(3x + 4y - 16 = 0\)

4.14. Distance from Point to Lines (Final Recap, Task 11)

Find the distance from point \(A(2, 1)\) to the following lines:

1. \(4x + 3y + 10 = 0\)

2. \(5x - 12y - 23 = 0\)

Click to see the solution

Key Concept: Distance from point \((x_0, y_0)\) to line \(Ax + By + C = 0\) is \(d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}\).

(a) For line \(4x + 3y + 10 = 0\) and point \((2, 1)\):

1. Apply the formula:

\[d = \frac{|4(2) + 3(1) + 10|}{\sqrt{4^2 + 3^2}} = \frac{|8 + 3 + 10|}{\sqrt{16 + 9}} = \frac{21}{\sqrt{25}} = \frac{21}{5}\]

(b) For line \(5x - 12y - 23 = 0\) and point \((2, 1)\):

1. Apply the formula:

\[d = \frac{|5(2) - 12(1) - 23|}{\sqrt{5^2 + (-12)^2}} = \frac{|10 - 12 - 23|}{\sqrt{25 + 144}} = \frac{|-25|}{\sqrt{169}} = \frac{25}{13}\]

Answer:

• 1. Distance is \(\frac{21}{5}\)
• 2. Distance is \(\frac{25}{13}\)

4.15. Equation of Plane Through Three Points (Final Recap, Task 12)

Find the equation of the plane passing through points \(M_1(2, 3, 1)\), \(M_2(3, 1, 4)\), \(M_3(2, 1, 5)\).

Click to see the solution

Key Concept: Find two vectors in the plane, compute their cross product to get the normal vector, then use the point-normal form.

1. Find vectors in the plane:

\[\vec{M_1M_2} = M_2 - M_1 = (3-2, 1-3, 4-1) = (1, -2, 3)\]

\[\vec{M_1M_3} = M_3 - M_1 = (2-2, 1-3, 5-1) = (0, -2, 4)\]

2. Compute the normal vector:

\[\vec{n} = \vec{M_1M_2} \times \vec{M_1M_3} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & 3 \\ 0 & -2 & 4 \end{vmatrix}\]

\[= \mathbf{i}((-2)(4) - (3)(-2)) - \mathbf{j}((1)(4) - (3)(0)) + \mathbf{k}((1)(-2) - (-2)(0))\]

\[= \mathbf{i}(-8 + 6) - \mathbf{j}(4) + \mathbf{k}(-2)\]

\[= (-2, -4, -2)\]

We can simplify by dividing by \(-2\): \(\vec{n} = (1, 2, 1)\)

3. Use point-normal form with \(M_1(2, 3, 1)\) and normal \((1, 2, 1)\):

\[1(x - 2) + 2(y - 3) + 1(z - 1) = 0\]

\[x - 2 + 2y - 6 + z - 1 = 0\]

\[x + 2y + z - 9 = 0\]

Answer: The equation of the plane is \(x + 2y + z - 9 = 0\).

4.16. Mutual Arrangement of Three Planes (Final Recap, Task 13)

Determine the mutual arrangement of the three planes: \(2x - 4y + 5z - 21 = 0\), \(x - 3z + 18 = 0\), \(6x + y + z - 30 = 0\).

Click to see the solution

Key Concept: Check if the normal vectors are linearly independent by computing their determinant.

1. Identify the normal vectors:

\[\vec{n}_1 = (2, -4, 5), \quad \vec{n}_2 = (1, 0, -3), \quad \vec{n}_3 = (6, 1, 1)\]

2. Compute the determinant:

\[\begin{vmatrix} 2 & -4 & 5 \\ 1 & 0 & -3 \\ 6 & 1 & 1 \end{vmatrix}\]

Expand along the second row:

\[= -1 \begin{vmatrix} -4 & 5 \\ 1 & 1 \end{vmatrix} + 0 - (-3) \begin{vmatrix} 2 & -4 \\ 6 & 1 \end{vmatrix}\]

\[= -1[(-4)(1) - (5)(1)] + 3[(2)(1) - (-4)(6)]\]

\[= -1(-4 - 5) + 3(2 + 24)\]

\[= -1(-9) + 3(26)\]

\[= 9 + 78 = 87 \neq 0\]

3. Conclusion: Since the determinant is non-zero, the normal vectors are linearly independent, meaning the planes are not parallel and intersect at a single point.

Answer: The three planes intersect at a single point.

4.17. Angle Between Lines in Space (Final Recap, Task 14)

Determine the angle between the lines \(\frac{x - 1}{3} = \frac{y + 2}{6} = \frac{z - 5}{2}\) and \(\frac{x}{2} = \frac{y - 3}{9} = \frac{z + 1}{6}\).

Click to see the solution

Key Concept: The angle between two lines equals the angle between their direction vectors.

1. Identify direction vectors:

\[\vec{v}_1 = (3, 6, 2), \quad \vec{v}_2 = (2, 9, 6)\]

2. Compute the dot product:

\[\vec{v}_1 \cdot \vec{v}_2 = (3)(2) + (6)(9) + (2)(6) = 6 + 54 + 12 = 72\]

3. Compute magnitudes:

\[|\vec{v}_1| = \sqrt{3^2 + 6^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7\]

\[|\vec{v}_2| = \sqrt{2^2 + 9^2 + 6^2} = \sqrt{4 + 81 + 36} = \sqrt{121} = 11\]

4. Find the angle:

\[\cos\theta = \frac{|\vec{v}_1 \cdot \vec{v}_2|}{|\vec{v}_1||\vec{v}_2|} = \frac{72}{7 \cdot 11} = \frac{72}{77}\]

\[\theta = \arccos\left(\frac{72}{77}\right)\]

Answer: The angle between the lines is \(\theta = \arccos\left(\frac{72}{77}\right)\) (approximately \(21.3°\)).

4.18. Line as Intersection of Planes (Final Recap, Task 15)

Derive canonical and parametric equations of the line represented as the intersection of two planes: \(\begin{cases} x - 2y + 4z = 0 \\ 3x - 2y + 5z = 0 \end{cases}\)

Click to see the solution

Key Concept: The direction vector of the line is the cross product of the normal vectors of the planes.

1. Find the direction vector:

Normal vectors: \(\vec{n}_1 = (1, -2, 4)\) and \(\vec{n}_2 = (3, -2, 5)\)

\[\vec{v} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & 4 \\ 3 & -2 & 5 \end{vmatrix}\]

\[= \mathbf{i}((-2)(5) - (4)(-2)) - \mathbf{j}((1)(5) - (4)(3)) + \mathbf{k}((1)(-2) - (-2)(3))\]

\[= \mathbf{i}(-10 + 8) - \mathbf{j}(5 - 12) + \mathbf{k}(-2 + 6)\]

\[= (-2, 7, 4)\]

2. Find a point on the line: Set \(z = 0\) and solve:

\[\begin{cases} x - 2y = 0 \\ 3x - 2y = 0 \end{cases}\]

Subtract first from second: \(2x = 0\), so \(x = 0\)

Substitute into first: \(0 - 2y = 0\), so \(y = 0\)

Point: \(M_0(0, 0, 0)\)

3. Write canonical form: With point \((0, 0, 0)\) and direction \((-2, 7, 4)\):

\[\frac{x - 0}{-2} = \frac{y - 0}{7} = \frac{z - 0}{4}\]

\[\frac{x}{-2} = \frac{y}{7} = \frac{z}{4}\]

4. Write parametric form: Let \(t\) be the parameter:

\[\begin{cases} x = -2t \\ y = 7t \\ z = 4t \end{cases}, \quad t \in \mathbb{R}\]

Answer:

• Canonical: \(\frac{x}{-2} = \frac{y}{7} = \frac{z}{4}\)
• Parametric: \(x = -2t\), \(y = 7t\), \(z = 4t\)

4.19. Canonical Equation of Ellipse (Part 1) (Final Recap, Task 16a)

Derive the canonical equation of the ellipse if the minor axis is 3 and eccentricity is \(\frac{\sqrt{2}}{2}\).

Click to see the solution

Key Concept: For an ellipse, \(b\) is the semi-minor axis, and the relationships are \(c^2 = a^2 - b^2\) and \(e = c/a\).

Given: \(b = 3\) (semi-minor axis) and \(e = \frac{\sqrt{2}}{2}\).

1. Use the eccentricity formula:

\[e = \frac{c}{a} = \frac{\sqrt{2}}{2}\]

\[e^2 = \frac{1}{2}\]

\[\frac{c^2}{a^2} = \frac{1}{2}\]

\[c^2 = \frac{a^2}{2}\]

2. Use the relationship \(c^2 = a^2 - b^2\):

\[\frac{a^2}{2} = a^2 - 9\]

\[\frac{a^2}{2} = a^2 - 9\]

\[a^2 = 2a^2 - 18\]

\[18 = a^2\]

\[a^2 = 18\]

3. Write the canonical equation:

\[\frac{x^2}{18} + \frac{y^2}{9} = 1\]

Answer: \(\frac{x^2}{18} + \frac{y^2}{9} = 1\)

4.20. Canonical Equation of Ellipse (Part 2) (Final Recap, Task 16b)

Derive the canonical equation of the ellipse if the distance between directrices equals 32 and \(e = 1/2\).

Click to see the solution

Key Concept: For an ellipse, the directrices are at \(x = \pm a/e\), so the distance between them is \(2a/e\).

1. Use the distance between directrices:

\[2 \cdot \frac{a}{e} = 32\]

\[\frac{a}{e} = 16\]

2. Substitute \(e = 1/2\):

\[\frac{a}{1/2} = 16\]

\[2a = 16\]

\[a = 8\]

3. Find \(c\) using \(e = c/a\):

\[c = ae = 8 \cdot \frac{1}{2} = 4\]

4. Find \(b\) using \(c^2 = a^2 - b^2\):

\[16 = 64 - b^2\]

\[b^2 = 48\]

5. Write the canonical equation:

\[\frac{x^2}{64} + \frac{y^2}{48} = 1\]

Answer: \(\frac{x^2}{64} + \frac{y^2}{48} = 1\)

4.21. Canonical Equation of Hyperbola (Final Recap, Task 17)

Derive the canonical equation of a hyperbola if the distance between directrices is \(288/13\) and the distance between foci is 26.

Click to see the solution

Key Concept: For a hyperbola, the distance between directrices is \(2a/e\) and the distance between foci is \(2c\).

1. Use given information:

Distance between directrices: \(2 \cdot \frac{a}{e} = \frac{288}{13}\)

Distance between foci: \(2c = 26\), so \(c = 13\)

2. From the directrices equation:

\[\frac{a}{e} = \frac{288}{26} = \frac{144}{13}\]

3. Also, \(e = c/a\), so:

\[\frac{a}{c/a} = \frac{144}{13}\]

\[\frac{a^2}{c} = \frac{144}{13}\]

\[a^2 = \frac{144c}{13} = \frac{144 \cdot 13}{13} = 144\]

\[a = 12\]

4. Find \(b\) using \(c^2 = a^2 + b^2\) (for hyperbola):

\[169 = 144 + b^2\]

\[b^2 = 25\]

5. Write the canonical equation:

\[\frac{x^2}{144} - \frac{y^2}{25} = 1\]

Answer: \(\frac{x^2}{144} - \frac{y^2}{25} = 1\)

4.22. Conic in Polar Coordinates (Final Recap, Task 18)

Identify the type of curve and derive the canonical form: \(\rho = \frac{1}{3 - 3\cos\varphi}\).

Click to see the solution

Key Concept: Convert from polar to Cartesian coordinates using \(\rho = \sqrt{x^2 + y^2}\) and \(x = \rho\cos\varphi\).

1. Rewrite the equation:

\[\rho(3 - 3\cos\varphi) = 1\]

\[3\rho - 3\rho\cos\varphi = 1\]

2. Substitute \(\rho = \sqrt{x^2 + y^2}\) and \(x = \rho\cos\varphi\):

\[3\sqrt{x^2 + y^2} - 3x = 1\]

\[3\sqrt{x^2 + y^2} = 1 + 3x\]

3. Square both sides:

\[9(x^2 + y^2) = (1 + 3x)^2\]

\[9x^2 + 9y^2 = 1 + 6x + 9x^2\]

\[9y^2 = 1 + 6x\]

4. Rearrange:

\[9y^2 = 6x + 1\]

\[y^2 = \frac{6x + 1}{9} = \frac{2}{3}\left(x + \frac{1}{6}\right)\]

5. Identify: This is a parabola with vertex at \((-1/6, 0)\) opening to the right.

Answer: Parabola with equation \(y^2 = \frac{2}{3}(x + \frac{1}{6})\)

4.23. Rotation to Canonical Form (Final Recap, Task 19)

Derive the canonical form and identify the type of conic section: \(25x^2 - 14xy + 25y^2 + 64x - 64y - 224 = 0\).

Click to see the solution

Key Concept: Use the discriminant to identify the conic type, then find the rotation angle to eliminate the \(xy\) term.

1. Compute the discriminant:

\[\Delta = B^2 - 4AC = (-14)^2 - 4(25)(25) = 196 - 2500 = -2304 < 0\]

Since \(\Delta < 0\), this is an ellipse.

2. Find the rotation angle: Use \(\cot 2\theta = \frac{A - C}{B}\):

\[\cot 2\theta = \frac{25 - 25}{-14} = \frac{0}{-14} = 0\]

\[2\theta = 90°\]

\[\theta = 45°\]

3. Apply rotation transformation for \(\theta = 45°\):

\[\begin{cases} x = \frac{1}{\sqrt{2}}(x' - y') \\ y = \frac{1}{\sqrt{2}}(x' + y') \end{cases}\]

4. Substitute into the original equation (this involves extensive algebra):

After substitution and simplification, we get:

\[32x'^2 + 18y'^2 + 64\sqrt{2}y' - 224 = 0\]

5. Complete the square for \(y'\):

\[32x'^2 + 18\left(y'^2 + \frac{64\sqrt{2}}{18}y'\right) = 224\]

\[32x'^2 + 18\left(y' + \frac{32\sqrt{2}}{18}\right)^2 - 18 \cdot \frac{2048}{324} = 224\]

After completing the square and simplifying (the solution shows):

\[32(x' + \sqrt{2})^2 + 18y'^2 = 288\]

6. Divide by 288:

\[\frac{(x' + \sqrt{2})^2}{9} + \frac{y'^2}{16} = 1\]

7. Final form: Setting \(x'' = x' + \sqrt{2}\) and \(y'' = y'\):

\[\frac{x''^2}{9} + \frac{y''^2}{16} = 1\]

Answer: Ellipse with canonical form \(\frac{x''^2}{9} + \frac{y''^2}{16} = 1\)

4.24. Identification of Quadric Surfaces (Final Recap, Task 20)

Identify the type of each quadric surface:

1. \(\frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{16} = 1\)

2. \(\frac{x^2}{4} - \frac{y^2}{9} - \frac{z^2}{16} = 1\)

3. \(z = \frac{x^2}{4} - \frac{y^2}{9}\)

4. \(\frac{x^2}{4} + \frac{y^2}{9} - \frac{z^2}{16} = 0\)

Click to see the solution

Key Concept: Identify quadric surfaces by examining the signs and arrangement of terms.

(1) \(\frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{16} = 1\)

All terms are positive and sum to 1. This is an ellipsoid.

(2) \(\frac{x^2}{4} - \frac{y^2}{9} - \frac{z^2}{16} = 1\)

One positive term equals 1 minus two other terms. The positive term can be arbitrarily large. This is a hyperboloid of two sheets (opening along the \(x\)-axis).

(3) \(z = \frac{x^2}{4} - \frac{y^2}{9}\)

One variable (z) equals a difference of squares. This is a hyperbolic paraboloid (saddle surface).

(4) \(\frac{x^2}{4} + \frac{y^2}{9} - \frac{z^2}{16} = 0\)

The equation equals zero (not 1). This is a cone with vertex at the origin.

Answer:

1. Ellipsoid
2. Hyperboloid of two sheets
3. Hyperbolic paraboloid
4. Cone

4.25. Intersection of Ellipsoid with Plane (Final Recap, Task 21)

Find the intersection of the ellipsoid \(x^2 + 4y^2 + 9z^2 = 36\) with the plane \(x = 2\).

Click to see the solution

Key Concept: Substitute the plane equation into the ellipsoid equation.

1. Substitute \(x = 2\) into the ellipsoid equation:

\[2^2 + 4y^2 + 9z^2 = 36\]

\[4 + 4y^2 + 9z^2 = 36\]

\[4y^2 + 9z^2 = 32\]

2. Divide by 32 to get standard form:

\[\frac{4y^2}{32} + \frac{9z^2}{32} = 1\]

\[\frac{y^2}{8} + \frac{z^2}{32/9} = 1\]

3. Identify: This is an ellipse in the plane \(x = 2\) with semi-axes \(\sqrt{8} = 2\sqrt{2}\) (along \(y\)) and \(\sqrt{32/9} = \frac{4\sqrt{2}}{3}\) (along \(z\)).

Answer: The intersection is an ellipse in the plane \(x = 2\) with equation \(\frac{y^2}{8} + \frac{z^2}{32/9} = 1\).

4.26. Angle Between Vectors (Additional Problems, Task 1)

Calculate the angle between the following vectors:

1. \(\mathbf{a} = (1, 1, 0)\) and \(\mathbf{b} = (0, -1, -1)\)

2. \(\mathbf{a} = (1/2, 1, 1)\) and \(\mathbf{b} = (-1, 1/2, 0)\)

Click to see the solution

Key Concept: Use \(\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\).

(a) For \(\mathbf{a} = (1, 1, 0)\) and \(\mathbf{b} = (0, -1, -1)\):

1. Compute dot product:

\[\mathbf{a} \cdot \mathbf{b} = (1)(0) + (1)(-1) + (0)(-1) = 0 - 1 + 0 = -1\]

2. Compute magnitudes:

\[|\mathbf{a}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}\]

\[|\mathbf{b}| = \sqrt{0^2 + (-1)^2 + (-1)^2} = \sqrt{2}\]

3. Find angle:

\[\cos\theta = \frac{-1}{\sqrt{2} \cdot \sqrt{2}} = \frac{-1}{2}\]

\[\theta = \arccos\left(-\frac{1}{2}\right) = 120°\]

(b) For \(\mathbf{a} = (1/2, 1, 1)\) and \(\mathbf{b} = (-1, 1/2, 0)\):

1. Compute dot product:

\[\mathbf{a} \cdot \mathbf{b} = \left(\frac{1}{2}\right)(-1) + (1)\left(\frac{1}{2}\right) + (1)(0) = -\frac{1}{2} + \frac{1}{2} + 0 = 0\]

2. Since the dot product is zero, the vectors are perpendicular:

\[\theta = 90°\]

Answer:

• 1. \(120°\)
• 2. \(90°\)

4.27. Projection and Reflection of Vectors (Additional Problems, Task 2)

Compute the orthogonal projection of vector \(\mathbf{a} = (2, 1, 2)\) onto \(\mathbf{b} = (1, 1, 1)\) and also compute the reflection of \(\mathbf{a}\) along vector \(\mathbf{b}\).

Click to see the solution

Key Concept: The projection is \(\text{proj}_{\mathbf{b}}\mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2}\mathbf{b}\). The reflection is \(\text{refl}_{\mathbf{b}}\mathbf{a} = 2\text{proj}_{\mathbf{b}}\mathbf{a} - \mathbf{a}\).

1. Compute dot product:

\[\mathbf{a} \cdot \mathbf{b} = (2)(1) + (1)(1) + (2)(1) = 2 + 1 + 2 = 5\]

2. Compute \(|\mathbf{b}|^2\):

\[|\mathbf{b}|^2 = 1^2 + 1^2 + 1^2 = 3\]

3. Find projection:

\[\text{proj}_{\mathbf{b}}\mathbf{a} = \frac{5}{3}(1, 1, 1) = \left(\frac{5}{3}, \frac{5}{3}, \frac{5}{3}\right)\]

4. Find reflection:

\[\text{refl}_{\mathbf{b}}\mathbf{a} = 2\text{proj}_{\mathbf{b}}\mathbf{a} - \mathbf{a}\]

\[= 2\left(\frac{5}{3}, \frac{5}{3}, \frac{5}{3}\right) - (2, 1, 2)\]

\[= \left(\frac{10}{3}, \frac{10}{3}, \frac{10}{3}\right) - (2, 1, 2)\]

\[= \left(\frac{10}{3} - 2, \frac{10}{3} - 1, \frac{10}{3} - 2\right)\]

\[= \left(\frac{4}{3}, \frac{7}{3}, \frac{4}{3}\right)\]

Answer:

• Projection: \(\left(\frac{5}{3}, \frac{5}{3}, \frac{5}{3}\right)\)
• Reflection: \(\left(\frac{4}{3}, \frac{7}{3}, \frac{4}{3}\right)\)

4.28. Determinant and Matrix Inverse (Additional Problems, Task 3)

Compute the determinant of the following matrix and also its inverse:

\[A = \begin{bmatrix} -1 & 0 & 1 \\ 2 & 0 & 0 \\ 2 & 1 & 3 \end{bmatrix}\]

Click to see the solution

Key Concept: Use cofactor expansion along a row or column with zeros for efficiency.

1. Compute determinant using cofactor expansion along the second row (which has two zeros):

\[\det(A) = -2 \begin{vmatrix} 0 & 1 \\ 1 & 3 \end{vmatrix} + 0 - 0\]

\[= -2[(0)(3) - (1)(1)] = -2(0 - 1) = -2(-1) = 2\]

2. Find the matrix of cofactors:

\[C_{11} = +\begin{vmatrix} 0 & 0 \\ 1 & 3 \end{vmatrix} = 0\]

\[C_{12} = -\begin{vmatrix} 2 & 0 \\ 2 & 3 \end{vmatrix} = -(6 - 0) = -6\]

\[C_{13} = +\begin{vmatrix} 2 & 0 \\ 2 & 1 \end{vmatrix} = 2 - 0 = 2\]

\[C_{21} = -\begin{vmatrix} 0 & 1 \\ 1 & 3 \end{vmatrix} = -(0 - 1) = 1\]

\[C_{22} = +\begin{vmatrix} -1 & 1 \\ 2 & 3 \end{vmatrix} = -3 - 2 = -5\]

\[C_{23} = -\begin{vmatrix} -1 & 0 \\ 2 & 1 \end{vmatrix} = -(-1 - 0) = 1\]

\[C_{31} = +\begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} = 0\]

\[C_{32} = -\begin{vmatrix} -1 & 1 \\ 2 & 0 \end{vmatrix} = -(0 - 2) = 2\]

\[C_{33} = +\begin{vmatrix} -1 & 0 \\ 2 & 0 \end{vmatrix} = 0\]

3. Form the adjugate matrix (transpose of cofactor matrix):

\[\text{adj}(A) = \begin{bmatrix} 0 & 1 & 0 \\ -6 & -5 & 2 \\ 2 & 1 & 0 \end{bmatrix}^T = \begin{bmatrix} 0 & -6 & 2 \\ 1 & -5 & 1 \\ 0 & 2 & 0 \end{bmatrix}\]

4. Compute inverse: \(A^{-1} = \frac{1}{\det(A)} \text{adj}(A)\)

\[A^{-1} = \frac{1}{2}\begin{bmatrix} 0 & 1 & 0 \\ -6 & -5 & 2 \\ 2 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1/2 & 0 \\ -3 & -5/2 & 1 \\ 1 & 1/2 & 0 \end{bmatrix}\]

Answer:

• Determinant: \(\det(A) = 2\)
• Inverse: \(A^{-1} = \begin{bmatrix} 0 & 1/2 & 0 \\ -3 & -5/2 & 1 \\ 1 & 1/2 & 0 \end{bmatrix}\)

4.29. Cross Product and Scalar Triple Product (Additional Problems, Task 4)

Given vectors \(\mathbf{a} = (1, 1, 0)\), \(\mathbf{b} = (0, 1, 1)\), \(\mathbf{c} = (-1, -3, 4)\), perform the following calculations:

1. Compute the cross product \(\mathbf{a} \times \mathbf{b}\)

2. Compute the scalar triple product \(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})\)

Then, interpret the geometric meaning of:

• The magnitude \(|\mathbf{a} \times \mathbf{b}|\)
• The absolute value \(|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|\)

Click to see the solution

(a) Cross product \(\mathbf{a} \times \mathbf{b}\):

1. Compute:

\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix}\]

\[= \mathbf{i}(1 \cdot 1 - 0 \cdot 1) - \mathbf{j}(1 \cdot 1 - 0 \cdot 0) + \mathbf{k}(1 \cdot 1 - 1 \cdot 0)\]

\[= \mathbf{i}(1) - \mathbf{j}(1) + \mathbf{k}(1) = (1, -1, 1)\]

2. Geometric meaning: \(|\mathbf{a} \times \mathbf{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}\) is the area of the parallelogram formed by \(\mathbf{a}\) and \(\mathbf{b}\).

(b) Scalar triple product:

1. First compute \(\mathbf{b} \times \mathbf{c}\):

\[\mathbf{b} \times \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 1 \\ -1 & -3 & 4 \end{vmatrix}\]

\[= \mathbf{i}(1 \cdot 4 - 1 \cdot (-3)) - \mathbf{j}(0 \cdot 4 - 1 \cdot (-1)) + \mathbf{k}(0 \cdot (-3) - 1 \cdot (-1))\]

\[= \mathbf{i}(4 + 3) - \mathbf{j}(0 + 1) + \mathbf{k}(0 + 1) = (7, -1, 1)\]

2. Compute the dot product:

\[\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = (1, 1, 0) \cdot (7, -1, 1) = 7 - 1 + 0 = 6\]

3. Geometric meaning: \(|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| = 6\) is the volume of the parallelepiped formed by the three vectors.

Answer:

• 1. \(\mathbf{a} \times \mathbf{b} = (1, -1, 1)\)
• 2. \(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 6\)
• \(|\mathbf{a} \times \mathbf{b}| = \sqrt{3}\) represents the parallelogram area
• \(|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| = 6\) represents the parallelepiped volume

4.30. Perpendicular Vector (Additional Problems, Task 5)

Calculate a vector that is perpendicular to \(\mathbf{a} = (1, -3, 2)\) and \(\mathbf{b} = (-2, 1, -5)\).

Click to see the solution

Key Concept: The cross product of two vectors is perpendicular to both.

1. Compute \(\mathbf{a} \times \mathbf{b}\):

\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -3 & 2 \\ -2 & 1 & -5 \end{vmatrix}\]

\[= \mathbf{i}((-3)(-5) - (2)(1)) - \mathbf{j}((1)(-5) - (2)(-2)) + \mathbf{k}((1)(1) - (-3)(-2))\]

\[= \mathbf{i}(15 - 2) - \mathbf{j}(-5 + 4) + \mathbf{k}(1 - 6)\]

\[= 13\mathbf{i} + \mathbf{j} - 5\mathbf{k} = (13, 1, -5)\]

Answer: \((13, 1, -5)\) (or any scalar multiple)

4.31. Ellipse: Graph and Eccentricity (Additional Problems, Task 5 continued)

Draw the graph of \(\frac{(x-1)^2}{9} + \frac{(y+1)^2}{4} = 1\) and calculate its eccentricity.

Click to see the solution

Key Concept: This is an ellipse centered at \((1, -1)\) with semi-major axis \(a = 3\) (along \(x\)) and semi-minor axis \(b = 2\) (along \(y\)).

1. Identify parameters:

• Center: \((h, k) = (1, -1)\)
• \(a = 3\) (semi-major axis, horizontal)
• \(b = 2\) (semi-minor axis, vertical)

2. Calculate \(c\):

\[c^2 = a^2 - b^2 = 9 - 4 = 5\]

\[c = \sqrt{5}\]

3. Calculate eccentricity:

\[e = \frac{c}{a} = \frac{\sqrt{5}}{3}\]

4. Graph description: The ellipse is centered at \((1, -1)\), extends 3 units horizontally and 2 units vertically from the center. Foci are at \((1 \pm \sqrt{5}, -1)\).

Answer: Eccentricity \(e = \frac{\sqrt{5}}{3} \approx 0.745\)

4.32. Locus of Points with Distance Ratio (Additional Problems, Task 6)

Find all points \(P(x, y)\) in the plane such that the distance from \(P\) to the point \((2, 0)\) is half the distance from \(P\) to the line \(x = 8\).

Click to see the solution

Key Concept: This is the definition of a conic section with focus at \((2, 0)\) and directrix \(x = 8\).

1. Set up the condition:

Distance from \(P(x, y)\) to \((2, 0)\): \(\sqrt{(x-2)^2 + y^2}\)

Distance from \(P(x, y)\) to line \(x = 8\): \(|x - 8|\)

Condition: \(\sqrt{(x-2)^2 + y^2} = \frac{1}{2}|x - 8|\)

2. Square both sides:

\[(x-2)^2 + y^2 = \frac{1}{4}(x-8)^2\]

3. Expand:

\[x^2 - 4x + 4 + y^2 = \frac{1}{4}(x^2 - 16x + 64)\]

\[x^2 - 4x + 4 + y^2 = \frac{1}{4}x^2 - 4x + 16\]

4. Simplify:

\[x^2 - \frac{1}{4}x^2 + y^2 = 16 - 4\]

\[\frac{3}{4}x^2 + y^2 = 12\]

\[\frac{x^2}{16} + \frac{y^2}{12} = 1\]

5. Identify: This is an ellipse centered at the origin with \(a = 4\) and \(b = 2\sqrt{3}\).

Answer: The locus is an ellipse with equation \(\frac{x^2}{16} + \frac{y^2}{12} = 1\)

4.33. Conic Rotation (Additional Problems, Task 7)

Determine the type of the following conic section and find a proper angle for rotating the axes to remove the \(xy\) term: \(2x^2 - 3xy + 2y^2 - 10 = 0\)

Click to see the solution

Key Concept: Use the discriminant to identify the type, then find the rotation angle.

1. Compute the discriminant:

\[\Delta = B^2 - 4AC = (-3)^2 - 4(2)(2) = 9 - 16 = -7 < 0\]

Since \(\Delta < 0\), this is an ellipse.

2. Find the rotation angle:

\[\cot 2\theta = \frac{A - C}{B} = \frac{2 - 2}{-3} = 0\]

\[2\theta = 90°\]

\[\theta = 45°\] or \(\theta = \frac{\pi}{4}\)

Answer:

• Type: Ellipse
• Rotation angle: \(\theta = 45°\) or \(\frac{\pi}{4}\)

4.34. Asymptotes of Hyperbola (Additional Problems, Task 8)

What are the asymptotes of the hyperbola \(\frac{(x-6)^2}{9} - \frac{(y+6)^2}{16} = 1\)?

Click to see the solution

Key Concept: For a hyperbola \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the asymptotes are \(y - k = \pm\frac{b}{a}(x - h)\).

1. Identify parameters:

• Center: \((h, k) = (6, -6)\)
• \(a = 3\)
• \(b = 4\)

2. Write asymptote equations:

\[y - (-6) = \pm\frac{4}{3}(x - 6)\]

\[y + 6 = \pm\frac{4}{3}(x - 6)\]

3. Two asymptotes:

\[y + 6 = \frac{4}{3}(x - 6) \quad \text{and} \quad y + 6 = -\frac{4}{3}(x - 6)\]

Answer:

• \(y + 6 = \frac{4}{3}(x - 6)\) or \(y = \frac{4}{3}x - 14\)
• \(y + 6 = -\frac{4}{3}(x - 6)\) or \(y = -\frac{4}{3}x + 2\)

4.35. Convert Conic to Standard Form (Additional Problems, Task 9)

Convert the conic section \(-9x^2 + 16y^2 - 72x - 96y - 144 = 0\) to the standard form, identify what type it is, and calculate its eccentricity.

Click to see the solution

Key Concept: Complete the square for both variables to convert to standard form.

1. Group terms:

\[-9x^2 - 72x + 16y^2 - 96y - 144 = 0\]

\[-9(x^2 + 8x) + 16(y^2 - 6y) = 144\]

2. Complete the square:

For \(x\): \(x^2 + 8x = (x + 4)^2 - 16\)

For \(y\): \(y^2 - 6y = (y - 3)^2 - 9\)

3. Substitute:

\[-9[(x + 4)^2 - 16] + 16[(y - 3)^2 - 9] = 144\]

\[-9(x + 4)^2 + 144 + 16(y - 3)^2 - 144 = 144\]

\[-9(x + 4)^2 + 16(y - 3)^2 = 144\]

4. Divide by 144:

\[\frac{16(y - 3)^2}{144} - \frac{9(x + 4)^2}{144} = 1\]

\[\frac{(y - 3)^2}{9} - \frac{(x + 4)^2}{16} = 1\]

5. Identify: This is a hyperbola centered at \((-4, 3)\) opening vertically, with \(a = 3\) and \(b = 4\).
6. Calculate eccentricity:

\[c^2 = a^2 + b^2 = 9 + 16 = 25\]

\[c = 5\]

\[e = \frac{c}{a} = \frac{5}{3}\]

Answer:

• Standard form: \(\frac{(y - 3)^2}{9} - \frac{(x + 4)^2}{16} = 1\)
• Type: Hyperbola
• Eccentricity: \(e = \frac{5}{3}\)

4.36. Plane Analysis (Additional Problems, Task 10)

Consider the plane \(\Pi_1\) given by \(2x - y + 3z = 5\).

1. Find a vector normal to \(\Pi_1\) and a point on it

2. Find the equation of a plane \(\Pi_2\) parallel to \(\Pi_1\) passing through \(P(1, 1, 1)\)

3. Find the distance from the point \(Q(4, 0, -2)\) to \(\Pi_1\)

4. Compute the angle between \(\Pi_1\) and \(\Pi_2\)

Click to see the solution

(a) Normal vector and a point:

1. Normal vector: From the equation \(2x - y + 3z = 5\), the normal vector is \(\vec{n} = (2, -1, 3)\).
2. Find a point: Set \(x = 0\) and \(y = 0\):

\[2(0) - 0 + 3z = 5\]

\[3z = 5\]

\[z = \frac{5}{3}\]

Point: \(\left(0, 0, \frac{5}{3}\right)\)

(b) Parallel plane through \(P(1, 1, 1)\):

1. Use same normal vector \((2, -1, 3)\):

\[2(x - 1) - 1(y - 1) + 3(z - 1) = 0\]

\[2x - 2 - y + 1 + 3z - 3 = 0\]

\[2x - y + 3z - 4 = 0\]

Or: \(2x - y + 3z = 4\)

(c) Distance from \(Q(4, 0, -2)\) to \(\Pi_1\):

1. Apply distance formula:

\[d = \frac{|2(4) - 1(0) + 3(-2) - 5|}{\sqrt{2^2 + (-1)^2 + 3^2}}\]

\[= \frac{|8 - 6 - 5|}{\sqrt{4 + 1 + 9}} = \frac{|-3|}{\sqrt{14}} = \frac{3}{\sqrt{14}} = \frac{3\sqrt{14}}{14}\]

(d) Angle between planes:

Since the planes are parallel, the angle is \(0°\).

Answer:

• 1. Normal: \((2, -1, 3)\); Point: \(\left(0, 0, \frac{5}{3}\right)\)
• 2. \(2x - y + 3z = 4\)
• 3. Distance: \(\frac{3\sqrt{14}}{14}\)
• 4. \(0°\) (parallel planes)

4.37. Relationship Between Lines in Space (Additional Problems, Task 11)

Let \(L_1: \frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z - 3}{4}\) and \(L_2: \begin{cases} x = 3 + t \\ y = 2 - t \\ z = 1 + 2t \end{cases}\). Determine whether \(L_1\) and \(L_2\) are parallel, intersecting, or skew.

Click to see the solution

Key Concept: Check if direction vectors are parallel, then check if lines intersect.

1. Identify direction vectors:

\(L_1\): \(\vec{v}_1 = (2, -1, 4)\)

\(L_2\): \(\vec{v}_2 = (1, -1, 2)\)

2. Check if parallel: Are \(\vec{v}_1\) and \(\vec{v}_2\) proportional?

\[\frac{2}{1} = 2, \quad \frac{-1}{-1} = 1, \quad \frac{4}{2} = 2\]

Not all equal, so not parallel.

3. Check for intersection: Convert \(L_1\) to parametric form:

\[\begin{cases} x = 1 + 2s \\ y = -1 - s \\ z = 3 + 4s \end{cases}\]

For \(L_2\): \(x = 3 + t\), \(y = 2 - t\), \(z = 1 + 2t\)

4. Set equal and solve:

From \(x\): \(1 + 2s = 3 + t \Rightarrow 2s - t = 2\) … (i)

From \(y\): \(-1 - s = 2 - t \Rightarrow -s + t = 3\) … (ii)

From (i) and (ii): Add them: \(s = 5\)

Substitute: \(t = -s + 3 = -5 + 3 = -2\)

5. Verify with \(z\):

\(L_1\): \(z = 3 + 4(5) = 23\)

\(L_2\): \(z = 1 + 2(-2) = -3\)

Since \(23 \neq -3\), the lines do not intersect.

Answer: The lines are skew (non-parallel and non-intersecting).

4.38. System of Linear Equations (Additional Problems, Task 12)

Consider the system:

\[\begin{cases} x + 2y - z = 1 \\ 2x + 5y + 3z = 4 \\ x + 4y + 7z = 7 \end{cases}\]

Write the augmented matrix and solve by Gaussian elimination and Cramer’s method.

Click to see the solution

Gaussian Elimination:

1. Write augmented matrix:

\[\left[\begin{array}{ccc|c} 1 & 2 & -1 & 1 \\ 2 & 5 & 3 & 4 \\ 1 & 4 & 7 & 7 \end{array}\right]\]

2. Eliminate first column below pivot:

\(R_2 - 2R_1\): \(\left[\begin{array}{ccc|c} 1 & 2 & -1 & 1 \\ 0 & 1 & 5 & 2 \\ 1 & 4 & 7 & 7 \end{array}\right]\)

\(R_3 - R_1\): \(\left[\begin{array}{ccc|c} 1 & 2 & -1 & 1 \\ 0 & 1 & 5 & 2 \\ 0 & 2 & 8 & 6 \end{array}\right]\)

3. Eliminate second column in \(R_3\):

\(R_3 - 2R_2\): \(\left[\begin{array}{ccc|c} 1 & 2 & -1 & 1 \\ 0 & 1 & 5 & 2 \\ 0 & 0 & -2 & 2 \end{array}\right]\)

4. Back substitution:

From \(R_3\): \(-2z = 2 \Rightarrow z = -1\)

From \(R_2\): \(y + 5(-1) = 2 \Rightarrow y = 7\)

From \(R_1\): \(x + 2(7) - (-1) = 1 \Rightarrow x + 15 = 1 \Rightarrow x = -14\)

Cramer’s Method:

1. Compute \(D\) (coefficient determinant):

\[D = \begin{vmatrix} 1 & 2 & -1 \\ 2 & 5 & 3 \\ 1 & 4 & 7 \end{vmatrix}\]

Expand along first row:

\[= 1 \begin{vmatrix} 5 & 3 \\ 4 & 7 \end{vmatrix} - 2 \begin{vmatrix} 2 & 3 \\ 1 & 7 \end{vmatrix} + (-1) \begin{vmatrix} 2 & 5 \\ 1 & 4 \end{vmatrix}\]

\[= 1(35 - 12) - 2(14 - 3) - 1(8 - 5)\]

\[= 23 - 22 - 3 = -2\]

2. Compute \(D_x\):

\[D_x = \begin{vmatrix} 1 & 2 & -1 \\ 4 & 5 & 3 \\ 7 & 4 & 7 \end{vmatrix} = 28\]

3. Compute \(D_y\):

\[D_y = \begin{vmatrix} 1 & 1 & -1 \\ 2 & 4 & 3 \\ 1 & 7 & 7 \end{vmatrix} = -14\]

4. Compute \(D_z\):

\[D_z = \begin{vmatrix} 1 & 2 & 1 \\ 2 & 5 & 4 \\ 1 & 4 & 7 \end{vmatrix} = 2\]

5. Solve:

\[x = \frac{D_x}{D} = \frac{28}{-2} = -14\]

\[y = \frac{D_y}{D} = \frac{-14}{-2} = 7\]

\[z = \frac{D_z}{D} = \frac{2}{-2} = -1\]

Answer: \((x, y, z) = (-14, 7, -1)\)

4.39. Non-Coplanar Vectors (Additional Problems, Task 13)

Show that \(\mathbf{a} = (1, 2, 3)\), \(\mathbf{b} = (2, -1, 1)\), and \(\mathbf{c} = (3, 1, -2)\) are not coplanar.

Click to see the solution

Key Concept: Three vectors are not coplanar if their scalar triple product is non-zero.

1. Compute the scalar triple product:

\[\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix} 1 & 2 & 3 \\ 2 & -1 & 1 \\ 3 & 1 & -2 \end{vmatrix}\]

2. Expand along first row:

\[= 1 \begin{vmatrix} -1 & 1 \\ 1 & -2 \end{vmatrix} - 2 \begin{vmatrix} 2 & 1 \\ 3 & -2 \end{vmatrix} + 3 \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix}\]

\[= 1(2 - 1) - 2(-4 - 3) + 3(2 + 3)\]

\[= 1 + 14 + 15 = 30 \neq 0\]

3. Conclusion: Since the scalar triple product is non-zero, the vectors are not coplanar.

Answer: The vectors are not coplanar (scalar triple product = 30).

4.40. Subspaces (Additional Problems, Task 14)

1. Does \(x + y + z = 0\) define a subspace in \(\mathbb{R}^3\)? How about \(x + 2y - z = 4\)? Why?

2. Are two vectors \(\mathbf{a} = (1, 2, 3)\) and \(\mathbf{b} = (3, 4, 1)\) linearly independent? How about \(\mathbf{a} = (1, 1)\), \(\mathbf{b} = (1, -1)\), \(\mathbf{c} = (1, 2)\)? Why?

Click to see the solution

(a) Subspace criteria:

For \(x + y + z = 0\):

1. Check zero vector: \((0, 0, 0)\) satisfies \(0 + 0 + 0 = 0\). ✓
2. Check closure under addition: If \(\vec{u}\) and \(\vec{v}\) satisfy the equation, does \(\vec{u} + \vec{v}\)?

Let \(\vec{u} = (u_1, u_2, u_3)\) with \(u_1 + u_2 + u_3 = 0\) and \(\vec{v} = (v_1, v_2, v_3)\) with \(v_1 + v_2 + v_3 = 0\).

Then \((u_1 + v_1) + (u_2 + v_2) + (u_3 + v_3) = (u_1 + u_2 + u_3) + (v_1 + v_2 + v_3) = 0 + 0 = 0\). ✓

3. Check closure under scalar multiplication: Similar reasoning. ✓

Yes, it’s a subspace (plane through the origin).

For \(x + 2y - z = 4\):

The zero vector \((0, 0, 0)\) doesn’t satisfy \(0 + 0 - 0 = 4\). Not a subspace (plane not through origin).

(b) Linear independence:

For \(\mathbf{a} = (1, 2, 3)\) and \(\mathbf{b} = (3, 4, 1)\) in \(\mathbb{R}^3\):

Check if one is a scalar multiple of the other:

\[\frac{3}{1} = 3, \quad \frac{4}{2} = 2, \quad \frac{1}{3} \neq 3\]

Not proportional, so linearly independent.

For \(\mathbf{a} = (1, 1)\), \(\mathbf{b} = (1, -1)\), \(\mathbf{c} = (1, 2)\) in \(\mathbb{R}^2\):

In \(\mathbb{R}^2\), any three vectors must be linearly dependent (dimension is 2).

To show explicitly, find scalars such that \(c_1\mathbf{a} + c_2\mathbf{b} + c_3\mathbf{c} = \mathbf{0}\):

\[c_1(1, 1) + c_2(1, -1) + c_3(1, 2) = (0, 0)\]

\[c_1 + c_2 + c_3 = 0\]

\[c_1 - c_2 + 2c_3 = 0\]

Solving: From the first equation, \(c_1 = -c_2 - c_3\). Substituting into the second: \(-c_2 - c_3 - c_2 + 2c_3 = -2c_2 + c_3 = 0\), so \(c_3 = 2c_2\). Then \(c_1 = -c_2 - 2c_2 = -3c_2\).

For \(c_2 = -1\): \(c_1 = 3\), \(c_3 = -2\). Verify: \(3(1,1) - 1(1,-1) - 2(1,2) = (3,3) + (-1,1) + (-2,-4) = (0,0)\).

Linearly dependent.

Answer:

• 1. \(x + y + z = 0\) is a subspace; \(x + 2y - z = 4\) is not (doesn’t contain origin)
• 2. First pair is independent; second set is dependent (more vectors than dimension)

4.41. Linear Transformation Matrix and Kernel (Additional Problems, Task 15)

Let \(T: \mathbb{R}^3 \to \mathbb{R}^2\) be defined by \(T(x, y, z) = (2x - y + z, x + 3y - 2z)\).

1. Find the matrix representation of \(T\) relative to standard bases

2. Find a basis for \(\text{ker}(T)\) and \(\text{im}(T)\)

3. Verify the Rank-Nullity Theorem for \(T\)

Click to see the solution

(a) Matrix representation:

1. Apply \(T\) to standard basis vectors:

\[T(1, 0, 0) = (2, 1)\]

\[T(0, 1, 0) = (-1, 3)\]

\[T(0, 0, 1) = (1, -2)\]

2. Form matrix with these as columns:

\[[T] = \begin{bmatrix} 2 & -1 & 1 \\ 1 & 3 & -2 \end{bmatrix}\]

(b) Basis for kernel and image:

Kernel: Solve \(T(x, y, z) = (0, 0)\):

\[\begin{cases} 2x - y + z = 0 \\ x + 3y - 2z = 0 \end{cases}\]

From second equation: \(x = -3y + 2z\)

Substitute into first: \(2(-3y + 2z) - y + z = 0\)

\[-6y + 4z - y + z = 0\]

\[-7y + 5z = 0\]

\[y = \frac{5z}{7}\]

\[x = -3 \cdot \frac{5z}{7} + 2z = -\frac{15z}{7} + \frac{14z}{7} = -\frac{z}{7}\]

Let \(z = 7\): \((x, y, z) = (-1, 5, 7)\)

Basis for \(\text{ker}(T)\): \(\{(-1, 5, 7)\}\)

Image: The columns of \([T]\) span the image. Since we have two linearly independent vectors in \(\mathbb{R}^2\):

Basis for \(\text{im}(T)\): \(\{(2, 1), (-1, 3)\}\)

(c) Rank-Nullity Theorem:

\[\dim(\text{ker}(T)) + \dim(\text{im}(T)) = \dim(\mathbb{R}^3)\]

\[1 + 2 = 3\] ✓

Answer:

• 1. \([T] = \begin{bmatrix} 2 & -1 & 1 \\ 1 & 3 & -2 \end{bmatrix}\)
• 2. Basis for \(\text{ker}(T)\): \(\{(-1, 5, 7)\}\); Basis for \(\text{im}(T)\): \(\{(2, 1), (-1, 3)\}\)
• 3. Verified: \(1 + 2 = 3\)

4.42. Triangle Area and Plane Equation (Additional Problems, Task 16)

Let \(P = (1, 2, -1)\), \(Q = (3, 0, 2)\), and \(R = (2, -1, 3)\).

1. Find the area of triangle \(PQR\)

2. Find the equation of the plane containing \(P\), \(Q\), and \(R\)

3. Find the distance from point \(S = (4, 1, 0)\) to this plane

Click to see the solution

(a) Area of triangle:

1. Find vectors:

\[\vec{PQ} = Q - P = (2, -2, 3)\]

\[\vec{PR} = R - P = (1, -3, 4)\]

2. Compute cross product:

\[\vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -2 & 3 \\ 1 & -3 & 4 \end{vmatrix}\]

\[= \mathbf{i}((-2)(4) - (3)(-3)) - \mathbf{j}((2)(4) - (3)(1)) + \mathbf{k}((2)(-3) - (-2)(1))\]

\[= \mathbf{i}(-8 + 9) - \mathbf{j}(8 - 3) + \mathbf{k}(-6 + 2)\]

\[= (1, -5, -4)\]

3. Find magnitude:

\[|\vec{PQ} \times \vec{PR}| = \sqrt{1^2 + (-5)^2 + (-4)^2} = \sqrt{1 + 25 + 16} = \sqrt{42}\]

4. Area is half the magnitude:

\[\text{Area} = \frac{1}{2}\sqrt{42}\]

(b) Plane equation:

1. Normal vector: \(\vec{n} = (1, -5, -4)\) from the cross product
2. Use point-normal form with \(P(1, 2, -1)\):

\[1(x - 1) - 5(y - 2) - 4(z - (-1)) = 0\]

\[x - 1 - 5y + 10 - 4z - 4 = 0\]

\[x - 5y - 4z + 5 = 0\]

(c) Distance from \(S(4, 1, 0)\) to plane:

\[d = \frac{|1(4) - 5(1) - 4(0) + 5|}{\sqrt{1^2 + (-5)^2 + (-4)^2}}\]

\[= \frac{|4 - 5 + 5|}{\sqrt{42}} = \frac{4}{\sqrt{42}} = \frac{4\sqrt{42}}{42} = \frac{2\sqrt{42}}{21}\]

Answer:

• 1. Area = \(\frac{\sqrt{42}}{2}\)
• 2. Plane: \(x - 5y - 4z + 5 = 0\)
• 3. Distance = \(\frac{4}{\sqrt{42}}\) or \(\frac{2\sqrt{42}}{21}\)

4.43. Identifying Subspaces (Additional Problems, Task 17)

Determine if the following sets are subspaces of \(\mathbb{R}^3\):

1. \(W_1 = \{(x, y, z) \mid x + 2y - z = 0\}\)

2. \(W_2 = \{(x, y, z) \mid x^2 + y^2 = z^2\}\)

3. \(W_3 = \{(x, y, z) \mid x \geq 0, y \geq 0, z \geq 0\}\)

4. \(W_4 = \{(x, y, z) \mid x + y = 1\}\)

Click to see the solution

Key Concept: Check if the set contains the zero vector and is closed under addition and scalar multiplication.

(a) \(W_1 = \{(x, y, z) \mid x + 2y - z = 0\}\):

• Zero vector: \((0, 0, 0)\) satisfies \(0 + 0 - 0 = 0\). ✓
• Closure properties: This is a plane through the origin. ✓

Yes, it’s a subspace.

(b) \(W_2 = \{(x, y, z) \mid x^2 + y^2 = z^2\}\) (cone):

• Zero vector: \((0, 0, 0)\) satisfies \(0 = 0\). ✓
• Closure under addition: Take \((1, 0, 1)\) and \((0, 1, 1)\), both in \(W_2\).

\((1, 0, 1) + (0, 1, 1) = (1, 1, 2)\)

Check: \(1^2 + 1^2 = 2 \neq 2^2 = 4\). ✗

Not a subspace (not closed under addition).

(c) \(W_3 = \{(x, y, z) \mid x \geq 0, y \geq 0, z \geq 0\}\) (first octant):

• Closure under scalar multiplication: Take \((1, 1, 1) \in W_3\) and scalar \(c = -1\).

\(-1 \cdot (1, 1, 1) = (-1, -1, -1) \notin W_3\). ✗

Not a subspace (not closed under scalar multiplication).

(d) \(W_4 = \{(x, y, z) \mid x + y = 1\}\) (plane not through origin):

• Zero vector: \((0, 0, 0)\) doesn’t satisfy \(0 + 0 = 1\). ✗

Not a subspace (doesn’t contain zero vector).

Answer:

• 1. Yes, subspace
• 2. No, not closed under addition
• 3. No, not closed under scalar multiplication
• 4. No, doesn’t contain zero vector

4.44. Matrix Representation of Linear Transformation (Additional Problems, Task 18)

Consider the linear transformation \(T: \mathbb{R}^2 \to \mathbb{R}^2\) defined by \(T(x, y) = (2x + y, x - 3y)\).

1. Find the matrix representation of \(T\) with respect to the standard basis

2. Find \(T(3, -2)\)

Click to see the solution

(a) Matrix representation:

1. Apply \(T\) to standard basis vectors:

\[T(1, 0) = (2 \cdot 1 + 0, 1 - 0) = (2, 1)\]

\[T(0, 1) = (2 \cdot 0 + 1, 0 - 3) = (1, -3)\]

2. Form matrix:

\[[T] = \begin{bmatrix} 2 & 1 \\ 1 & -3 \end{bmatrix}\]

(b) Compute \(T(3, -2)\):

1. Direct calculation:

\[T(3, -2) = (2(3) + (-2), 3 - 3(-2)) = (6 - 2, 3 + 6) = (4, 9)\]

Answer:

• 1. \([T] = \begin{bmatrix} 2 & 1 \\ 1 & -3 \end{bmatrix}\)
• 2. \(T(3, -2) = (4, 9)\)

4.45. Change of Basis (Additional Problems, Task 19)

Let \(V = \mathbb{R}^2\) with the standard basis \(E = \{\mathbf{e}_1 = (1, 0), \mathbf{e}_2 = (0, 1)\}\). Consider the basis \(B = \{\mathbf{v}_1 = (2, 1), \mathbf{v}_2 = (1, 3)\}\).

1. Find the transition matrix \(P_{E \leftarrow B}\) from basis \(B\) to basis \(E\)

2. Find the transition matrix \(P_{B \leftarrow E}\) from basis \(E\) to basis \(B\)

3. Verify that \(P_{B \leftarrow E} = P^{-1}_{E \leftarrow B}\)

4. Let \(\mathbf{x} = (5, 4)\) in the standard basis. Find its coordinates \([\mathbf{x}]_B\) in basis \(B\)

Click to see the solution

(a) Transition matrix \(P_{E \leftarrow B}\):

The columns are the basis vectors of \(B\) expressed in basis \(E\):

\[P_{E \leftarrow B} = \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}\]

(b) Transition matrix \(P_{B \leftarrow E}\):

This is the inverse of \(P_{E \leftarrow B}\):

\[\det(P_{E \leftarrow B}) = 2(3) - 1(1) = 6 - 1 = 5\]

\[P_{B \leftarrow E} = \frac{1}{5}\begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix}\]

(c) Verification:

\[P_{E \leftarrow B} \cdot P_{B \leftarrow E} = \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix} \cdot \frac{1}{5}\begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix}\]

\[= \frac{1}{5}\begin{bmatrix} 6-1 & -2+2 \\ 3-3 & -1+6 \end{bmatrix} = \frac{1}{5}\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\] ✓

(d) Coordinates in basis \(B\):

\[[\mathbf{x}]_B = P_{B \leftarrow E}[\mathbf{x}]_E = \frac{1}{5}\begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 5 \\ 4 \end{bmatrix}\]

\[= \frac{1}{5}\begin{bmatrix} 15-4 \\ -5+8 \end{bmatrix} = \frac{1}{5}\begin{bmatrix} 11 \\ 3 \end{bmatrix} = \begin{bmatrix} 11/5 \\ 3/5 \end{bmatrix}\]

Answer:

• 1. \(P_{E \leftarrow B} = \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}\)
• 2. \(P_{B \leftarrow E} = \frac{1}{5}\begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix}\)
• 3. Verified
• 4. \([\mathbf{x}]_B = \left(\frac{11}{5}, \frac{3}{5}\right)\)

4.46. Identification of Quadric Surfaces II (Additional Problems, Task 20)

Identify the following quadric surfaces:

1. \(\frac{x^2}{9} + \frac{y^2}{4} + \frac{z^2}{16} = 1\)

2. \(\frac{x^2}{4} - \frac{y^2}{9} + \frac{z^2}{1} = 1\)

3. \(z = x^2 + y^2\)

4. \(z = x^2 - y^2\)

Click to see the solution

Key Concept: Classify based on the signs and structure of the equation.

(a) \(\frac{x^2}{9} + \frac{y^2}{4} + \frac{z^2}{16} = 1\)

All terms positive, sum to 1: Ellipsoid

(b) \(\frac{x^2}{4} - \frac{y^2}{9} + \frac{z^2}{1} = 1\)

Mixed signs, equation equals 1: Hyperboloid of one sheet (two positive terms)

(c) \(z = x^2 + y^2\)

One variable linear, others squared with same sign: Elliptic paraboloid (circular cross-sections)

(d) \(z = x^2 - y^2\)

One variable linear, others squared with opposite signs: Hyperbolic paraboloid (saddle surface)

Answer:

• 1. Ellipsoid
• 2. Hyperboloid of one sheet
• 3. Elliptic paraboloid (circular paraboloid)
• 4. Hyperbolic paraboloid