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Scalar multiplication of vectors

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Difficulty: 1 – 5

52774312

The point \(P(4, -2, 5)\) is translated by \(\mathbf{v}=\begin{pmatrix}-3\\6\\1\end{pmatrix}\) to point \(Q\). a) Find \(Q\). b) Starting at \(Q\), translate by \(2\mathbf{v}\) to point \(R\). Find \(R\).

Hints

- A translation adds a vector to a point's coordinates. - Multiply every component to find \(2\mathbf{v}\). - Apply the second translation to \(Q\), not to \(P\). - The total translation from \(P\) to \(R\) is \(3\mathbf{v}\).

Solution

1. Add the translation vector to \(P\): \(Q=P+\mathbf{v}=(4, -2, 5)+(-3, 6, 1)=(1, 4, 6)\). 2. \(2\mathbf{v}=\begin{pmatrix}-6\\12\\2\end{pmatrix}\). 3. Add this vector to \(Q\): \(R=Q+2\mathbf{v}=(1, 4, 6)+(-6, 12, 2)=(-5, 16, 8)\).

Answer

a) \(Q=(1, 4, 6)\) b) \(R=(-5, 16, 8)\)

52775412

The vector \(\mathbf{v}=\begin{pmatrix}1\\-3\\\frac{3}{4}\end{pmatrix}\) is given. Find \(\mathbf{w}=-4\mathbf{v}\) and \(\lVert\mathbf{w}\rVert\).

Hints

- Scalar multiplication affects every component. - Use the magnitude formula after finding the new vector. - Squared components are nonnegative even when the component is negative.

Solution

1. Multiply each component by \(-4\): \(\mathbf{w}=\begin{pmatrix}-4\\12\\-3\end{pmatrix}\). 2. \(\lVert\mathbf{w}\rVert=\sqrt{(-4)^2+12^2+(-3)^2}=\sqrt{169}=13\).

Answer

\(\mathbf{w}=\begin{pmatrix}-4\\12\\-3\end{pmatrix}\), \(\lVert\mathbf{w}\rVert=13\)

52775912

For each expression, find the resulting vector and its magnitude. a) \(\mathbf{a}=2.5\begin{pmatrix}2\\-4\\4\end{pmatrix}\) b) \(\mathbf{b}=1.8\begin{pmatrix}0\\4\\-3\end{pmatrix}-0.8\begin{pmatrix}0\\4\\-3\end{pmatrix}\)

Hints

- Multiply every component by the scalar. - In part b), combine the scalar coefficients first. - Use the three-dimensional magnitude formula.

Solution

1. In part a), \(\mathbf{a}=\begin{pmatrix}5\\-10\\10\end{pmatrix}\), so \(\lVert\mathbf{a}\rVert=\sqrt{5^2+(-10)^2+10^2}=15\). 2. In part b), factor the common vector: \(\mathbf{b}=(1.8-0.8)\begin{pmatrix}0\\4\\-3\end{pmatrix} =\begin{pmatrix}0\\4\\-3\end{pmatrix}\). Thus, \(\lVert\mathbf{b}\rVert=\sqrt{4^2+(-3)^2}=5\).

Answer

a) \(\mathbf{a}=\begin{pmatrix}5\\-10\\10\end{pmatrix}\), \(\lVert\mathbf{a}\rVert=15\) b) \(\mathbf{b}=\begin{pmatrix}0\\4\\-3\end{pmatrix}\), \(\lVert\mathbf{b}\rVert=5\)

52776112

Write \(\mathbf{a}=\begin{pmatrix}\frac{5}{6}\\-0.25\\1\end{pmatrix}\) as the product of a real number and a vector with integer components.

Hints

- Rewrite the decimal as a fraction. - Find a common denominator for all components. - Factor the reciprocal of that denominator out of the vector.

Solution

1. Rewrite \(-0.25\) as \(-\frac{1}{4}\). 2. The least common denominator of \(6\) and \(4\) is \(12\). 3. Write the components with denominator \(12\): \(\frac{5}{6}=\frac{10}{12}\), \(-\frac{1}{4}=-\frac{3}{12}\), and \(1=\frac{12}{12}\). 4. Factor out \(\frac{1}{12}\): \(\mathbf{a}=\frac{1}{12}\begin{pmatrix}10\\-3\\12\end{pmatrix}\).

Answer

\(\mathbf{a}=\frac{1}{12}\begin{pmatrix}10\\-3\\12\end{pmatrix}\)

52776312

For each vector equation, determine whether a real number \(k\) exists. Give \(k\), or explain why no such value exists. a) \(\begin{pmatrix}12\\-18\\30\end{pmatrix}=k\begin{pmatrix}-4\\6\\-10\end{pmatrix}\) b) \(\begin{pmatrix}\frac{2}{3}\\1\\\frac{5}{6}\end{pmatrix}=k\begin{pmatrix}4\\6\\4\end{pmatrix}\)

Hints

- Treat corresponding components as separate equations. - One scalar must satisfy every component equation. - Compare the values of \(k\) obtained from each component. - Simplify fractions before deciding whether the values agree.

Solution

1. In part a), compare corresponding components: \(12=-4k\), \(-18=6k\), and \(30=-10k\). Each equation gives \(k=-3\), so the vector equation is satisfied. 2. In part b), the first component gives \(k=\frac{1}{6}\), and the second also gives \(k=\frac{1}{6}\). The third gives \(k=\frac{5}{24}\). 3. Since one value of \(k\) does not satisfy all three components, no such real number exists in part b).

Answer

a) \(k=-3\) b) No such real number exists.

53028112

The vectors \(\mathbf{a}=\begin{pmatrix}3\\-1\\2\end{pmatrix}\), \(\mathbf{b}=\begin{pmatrix}-9\\3\\-6\end{pmatrix}\), and \(\mathbf{c}=\begin{pmatrix}1.5\\-0.5\\1\end{pmatrix}\) are given. a) Find real numbers \(r\) and \(s\) such that \(\mathbf{b}=r\mathbf{a}\) and \(\mathbf{c}=s\mathbf{a}\). b) The vector \(\mathbf{d}=\begin{pmatrix}x\\4\\z\end{pmatrix}\) is also a scalar multiple of \(\mathbf{a}\). Find \(x\) and \(z\).

Hints

- Compare corresponding components. - One scale factor must work for every component. - Use the known component of \(\mathbf{d}\) to find its scale factor. - Apply that factor to the remaining components.

Solution

1. Comparing components of \(\mathbf{b}=r\mathbf{a}\) gives \(r=-3\). 2. Comparing components of \(\mathbf{c}=s\mathbf{a}\) gives \(s=0.5\). 3. Write \(\mathbf{d}=k\mathbf{a}\). From the second component, \(4=-k\), so \(k=-4\). 4. Therefore, \(x=3\cdot(-4)=-12\) and \(z=2\cdot(-4)=-8\).

Answer

a) \(r=-3\), \(s=0.5\) b) \(x=-12\), \(z=-8\)

53028712

Consider \(\mathbf{a}=\begin{pmatrix}7\\-2.5\\k\end{pmatrix}\), where \(k\) is real. a) Write the opposite vector \(-\mathbf{a}\). b) Find \(k\) so that the sum of the components of \(-\mathbf{a}\) is \(0\). c) State the general rule for finding the opposite of \(\mathbf{v}=\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}\).

Hints

- Multiplying by \(-1\) changes every sign. - Add the three components and set the sum equal to \(0\). - A vector plus its opposite equals \(\mathbf{0}\).

Solution

1. Multiply each component by \(-1\): \(-\mathbf{a}=\begin{pmatrix}-7\\2.5\\-k\end{pmatrix}\). 2. Set the component sum equal to \(0\): \(-7+2.5-k=0\). 3. This simplifies to \(-4.5-k=0\), so \(k=-4.5\). 4. In general, \(-\mathbf{v}=\begin{pmatrix}-v_1\\-v_2\\-v_3\end{pmatrix}\).

Answer

a) \(-\mathbf{a}=\begin{pmatrix}-7\\2.5\\-k\end{pmatrix}\) b) \(k=-4.5\) c) Negate every component.

53028912

A triangle has vertices \(A(3, 0, -1)\), \(B(1, 4, 2)\), and \(C(-2, 2, 5)\). It is translated by a vector \(\mathbf{v}\) that has twice the magnitude of \(\mathbf{u}=\begin{pmatrix}1\\-2\\0\end{pmatrix}\) and points in the same direction. a) Find \(\mathbf{v}\). b) Find the image vertices \(A'\), \(B'\), and \(C'\).

Hints

- A positive scalar multiple preserves direction. - Doubling a vector doubles every component. - A translation adds the same vector to every point.

Solution

1. Since \(\mathbf{v}\) points in the same direction and has twice the magnitude, \(\mathbf{v}=2\mathbf{u}=\begin{pmatrix}2\\-4\\0\end{pmatrix}\). 2. Add \(\mathbf{v}\) to each vertex: \(A'=(3, 0, -1)+(2, -4, 0)=(5, -4, -1)\), \(B'=(1, 4, 2)+(2, -4, 0)=(3, 0, 2)\), and \(C'=(-2, 2, 5)+(2, -4, 0)=(0, -2, 5)\).

Answer

a) \(\mathbf{v}=\begin{pmatrix}2\\-4\\0\end{pmatrix}\) b) \(A'=(5, -4, -1)\), \(B'=(3, 0, 2)\), \(C'=(0, -2, 5)\)

53029412

A translation maps \(P(2, 0, -5)\) to \(Q(5, -3, 1)\). 1. Find the translation vector \(\mathbf{v}\). 2. Starting at \(P\), apply the translation \(3\mathbf{v}\) to find \(R\).

Hints

- Find the translation vector as \(Q-P\). - Multiply every component by \(3\). - Add the resulting displacement to \(P\).

Solution

1. \(\mathbf{v}=Q-P=\begin{pmatrix}3\\-3\\6\end{pmatrix}\). 2. \(3\mathbf{v}=\begin{pmatrix}9\\-9\\18\end{pmatrix}\). 3. \(R=P+3\mathbf{v}=(2, 0, -5)+(9, -9, 18)=(11, -9, 13)\).

Answer

1. \(\mathbf{v}=\begin{pmatrix}3\\-3\\6\end{pmatrix}\) 2. \(R=(11, -9, 13)\)

53037912

Points \(A(9, -6, 12)\), \(B(-3, 2, -4)\), \(C(6, -4, 10)\), and \(D\left(-\frac{3}{2}, 1, -2\right)\) are given. 1. Determine which points lie on the same line through the origin. 2. Write a parametric equation of that line.

Hints

- What relationship must position vectors have for points on one line through the origin? - Compare corresponding coordinate ratios. - All ratios must agree for a scalar multiple. - A line through the origin has the zero vector as its position vector.

Solution

1. Points on the same line through the origin have position vectors that are scalar multiples of one direction vector. 2. Using \(\begin{pmatrix}3\\-2\\4\end{pmatrix}\), we have \(\overrightarrow{OA}=3\begin{pmatrix}3\\-2\\4\end{pmatrix}\), \(\overrightarrow{OB}=-\begin{pmatrix}3\\-2\\4\end{pmatrix}\), and \(\overrightarrow{OD}=-\frac{1}{2}\begin{pmatrix}3\\-2\\4\end{pmatrix}\). 3. For \(C\), the first two coordinate ratios are \(2\), but the third is \(\frac{5}{2}\), so \(C\) is not on the line. 4. Therefore, \(A\), \(B\), and \(D\) lie on \(g:\vec{x}=t\begin{pmatrix}3\\-2\\4\end{pmatrix}\).

Answer

1. Points \(A\), \(B\), and \(D\) lie on the same line through the origin; \(C\) does not. 2. \(g:\vec{x}=t\begin{pmatrix}3\\-2\\4\end{pmatrix}\), where \(t\in\mathbb{R}\)

53039212

Classify quadrilateral \(ABCD\) with vertices \(A(1, 0, 2)\), \(B(5, 2, 4)\), \(C(4, 3, 5)\), and \(D(2, 2, 4)\) as a trapezoid, parallelogram, or rhombus. Use the inclusive definition of a trapezoid.

Hints

- Compute vectors for a pair of opposite sides. - Parallel vectors are scalar multiples. - A rhombus must first be a parallelogram.

Solution

1. \(\overrightarrow{AB}=(4, 2, 2)\) and \(\overrightarrow{DC}=C-D=(2, 1, 1)\). 2. Since \(\overrightarrow{AB}=2\overrightarrow{DC}\), sides \(\overline{AB}\) and \(\overline{DC}\) are parallel. Therefore, \(ABCD\) is a trapezoid. 3. The opposite side vectors are not equal, so \(ABCD\) is not a parallelogram. 4. Since every rhombus is a parallelogram, \(ABCD\) is not a rhombus.

Answer

\(ABCD\) is a trapezoid, but it is not a parallelogram or a rhombus.

53042712

Let \(\vec{u}=\begin{pmatrix}3\\-4\\12\end{pmatrix}\). a) Find \(\|\vec{u}\|\). b) Let \(\vec{v}=-2\vec{u}\). Find \(\|\vec{v}\|\) in two ways: 1. Compute \(\vec{v}\) first. 2. Use \(\|k\vec{a}\|=|k|\|\vec{a}\|\). c) Explain how the factor \(-2\) affects direction and magnitude.

Hints

- Compute the original magnitude first. - Scalar multiplication changes every component. - Magnitude is multiplied by the absolute value of the scalar.

Solution

1. \(\|\vec{u}\|=\sqrt{3^2+(-4)^2+12^2}=\sqrt{169}=13\). 2. \(\vec{v}=\begin{pmatrix}-6\\8\\-24\end{pmatrix}\), so \(\|\vec{v}\|=\sqrt{36+64+576}=26\). 3. Alternatively, \(\|\vec{v}\|=|-2|\|\vec{u}\|=2\cdot13=26\). 4. The negative sign reverses the direction, and the absolute value \(2\) doubles the magnitude.

Answer

a) \(13\) b) \(\vec{v}=\begin{pmatrix}-6\\8\\-24\end{pmatrix}\) and \(\|\vec{v}\|=26\) by both methods. c) The direction reverses and the magnitude doubles.

53042812

Let \(\vec{a}=\begin{pmatrix}4\\0\\-3\end{pmatrix}\). a) Find \(\|\vec{a}\|\). b) A vector \(\vec{b}=r\vec{a}\) has magnitude \(20\). Find all possible values of \(r\). c) Interpret the two values geometrically.

Hints

- Use \(\|r\vec{a}\|=|r|\|\vec{a}\|\). - Solve the absolute-value equation. - Interpret the sign of the scalar separately from its magnitude.

Solution

1. \(\|\vec{a}\|=\sqrt{4^2+0^2+(-3)^2}=5\). 2. Since \(\|\vec{b}\|=|r|\|\vec{a}\|\), \(20=5|r|\), so \(|r|=4\). 3. Therefore, \(r=4\) or \(r=-4\). 4. For \(r=4\), \(\vec{b}\) points in the same direction as \(\vec{a}\). For \(r=-4\), it points in the opposite direction. In both cases, its magnitude is four times as large.

Answer

a) \(5\) b) \(r=4\) or \(r=-4\) c) The vectors have opposite possible directions but the same magnitude \(20\).

53061312

Consider the planes \(E_1: 4x-6y+2z=10\), \(E_2: -2x+3y-z=-5\), \(E_3: 2x-3y+z=7\). a) Determine the relationship between \(E_1\) and \(E_2\). b) Determine the relationship between \(E_2\) and \(E_3\). c) Explain how proportional normal vectors and constants distinguish identical planes from distinct parallel planes.

Hints

- Compare the coefficients of \(x\), \(y\), and \(z\). - Find the scale factor between each pair of normal vectors. - Apply the same scale factor to the constant term. - Decide whether the full equations are multiples or only the normal vectors are.

Solution

1. a) The normal vectors are \(\vec{n}_1=\langle 4,-6,2\rangle\) and \(\vec{n}_2=\langle -2,3,-1\rangle\). Since \(\vec{n}_1=-2\vec{n}_2\), the planes are parallel or identical. 2. The constants have the same scale factor because \(10=-2(-5)\). Thus the entire equation of \(E_1\) is \(-2\) times the equation of \(E_2\), so the planes are identical. 3. b) The normal vectors of \(E_2\) and \(E_3\) satisfy \(\vec{n}_2=-\vec{n}_3\), so these planes are parallel or identical. 4. Multiplying the equation of \(E_3\) by \(-1\) gives \(-2x+3y-z=-7\), not \(-5\). Therefore \(E_2\) and \(E_3\) are distinct and strictly parallel. 5. c) Proportional normal vectors mean the planes have the same orientation. If the constants have the same proportionality factor, the equations describe the same plane; otherwise, they describe distinct parallel planes.

Answer

a) \(E_1\) and \(E_2\) are identical. b) \(E_2\) and \(E_3\) are strictly parallel. c) Proportional normals give parallel orientation; matching proportional constants give identical planes.

53061612

Planes \(E_1\) and \(E_2\) are given by \(E_1:3x-6y+9z=12\), \(E_2:-x+2y-3z=k\), where \(k\in\mathbb{R}\). a) Find the value of \(k\) for which the planes are identical. b) Describe the relationship between the planes when \(k=0\).

Hints

- Compare the normal vectors as scalar multiples. - Normalize both plane equations. - Equal left sides with equal constants describe the same plane; unequal constants describe strictly parallel planes.

Solution

1. The normal vectors are \(\vec{n}_1=\langle 3,-6,9\rangle\) and \(\vec{n}_2=\langle -1,2,-3\rangle\). 2. Since \(\vec{n}_1=-3\vec{n}_2\), the planes are either identical or strictly parallel. 3. a) Divide the equation of \(E_1\) by \(3\): \(x-2y+3z=4\). 4. Multiply the equation of \(E_2\) by \(-1\): \(x-2y+3z=-k\). 5. The planes are identical when \(-k=4\), so \(k=-4\). 6. b) When \(k=0\), the normalized equations have the same left side but different right sides, \(4\) and \(0\). 7. Therefore, the planes are strictly parallel.

Answer

a) \(k=-4\) b) For \(k=0\), the planes are strictly parallel.

53062212

Plane \(E\) is given by \(3x+4z=12\). a) Plane \(F\) is given by \(-1.5x-2z=-6\). Determine the relationship between \(E\) and \(F\). b) Find an equation of a plane \(G\) that is parallel to \(E\) and passes through the origin. c) Describe the family of all planes parallel to \(E\).

Hints

- Multiply one entire equation by a constant and compare it with the other. - Parallel planes have parallel normal vectors. - Substitute the origin into the general parallel-plane equation. - The constant controls the plane’s offset, not its orientation.

Solution

1. a) Multiplying the equation of \(F\) by \(-2\) gives \(3x+4z=12\), which is exactly the equation of \(E\). Therefore the planes are identical. 2. b) A plane parallel to \(E\) can use the normal vector \(\langle 3,0,4\rangle\), so write \(3x+4z=d\). 3. The origin gives \(d=0\), so \(G:3x+4z=0\). 4. c) The family of all parallel planes is \(3x+4z=d\), where \(d\in\mathbb{R}\).

Answer

a) \(E\) and \(F\) are identical. b) \(G:3x+4z=0\) c) \(3x+4z=d\), where \(d\in\mathbb{R}\)

53062812

Plane \(E\) and the family of planes \(F_t\) are given by \(E:5x-2y+z=4\), \(F_t:10x-4y+2z=t\), where \(t\in\mathbb{R}\). a) Show that every plane \(F_t\) is parallel to or identical with \(E\). b) Find \(t\) so that \(F_t\) contains \(P(1, 1, 1)\). c) For that value of \(t\), determine the relationship between \(F_t\) and \(E\).

Hints

- Compare the normal vectors as scalar multiples. - Substitute the given point into the family equation. - Normalize the resulting plane equation.

Solution

1. a) The normal vectors are \(\vec{n}_E=\langle 5,-2,1\rangle\) and \(\vec{n}_{F_t}=\langle 10,-4,2\rangle\). 2. Since \(\vec{n}_{F_t}=2\vec{n}_E\), every \(F_t\) is parallel to or identical with \(E\). 3. b) Substitute \(P(1, 1, 1)\) into \(F_t\): \(10-4+2=t\), so \(t=8\). 4. c) Dividing \(F_8:10x-4y+2z=8\) by \(2\) gives \(5x-2y+z=4\), exactly the equation of \(E\). 5. Therefore, \(F_8\) and \(E\) are identical.

Answer

a) \(\vec{n}_{F_t}=2\vec{n}_E\), so every \(F_t\) is parallel to or identical with \(E\). b) \(t=8\) c) \(F_8\) and \(E\) are identical.

52540912

Find all real values of \(a\) for which the vectors \(\vec{u}=\begin{pmatrix}a\\2\\2\end{pmatrix}\) and \(\vec{v}=\begin{pmatrix}8\\a\\a\end{pmatrix}\) are linearly dependent.

Hints

- When are two vectors linearly dependent? - Set one vector equal to a scalar multiple of the other. - Use the repeated second and third components to express the scale factor. - Remember that a quadratic equation may have two real solutions.

Solution

1. Two nonzero vectors are linearly dependent when one is a scalar multiple of the other. Let \(\vec{v}=k\vec{u}\). 2. The coordinate equations are \(8=ka\) and \(a=2k\). Thus \(k=\frac{a}{2}\). 3. Substitute into the first equation: \(8=\frac{a}{2}a\), so \(a^2=16\). 4. Therefore, \(a=4\) or \(a=-4\). For these values, the scale factors are \(2\) and \(-2\), respectively.

Answer

\(a=4\) or \(a=-4\)

52772112

The points \(P(2, -3, 1)\) and \(Q(-1, 1, 1)\) are given. a) Find \(\overrightarrow{PQ}\) and its magnitude. b) A point \(R\) is defined by \(\overrightarrow{QR}=2\overrightarrow{PQ}\). Find the coordinates of \(R\).

Hints

- Compute a displacement vector as terminal point minus initial point. - Use the magnitude formula for a three-dimensional vector. - Scalar multiplication multiplies every component. - Add the displacement vector to the starting point \(Q\).

Solution

1. \(\overrightarrow{PQ}=Q-P=\begin{pmatrix}-3\\4\\0\end{pmatrix}\). 2. Its magnitude is \(\lVert\overrightarrow{PQ}\rVert=\sqrt{(-3)^2+4^2+0^2}=5\). 3. \(\overrightarrow{QR}=2\overrightarrow{PQ}=\begin{pmatrix}-6\\8\\0\end{pmatrix}\). 4. Add this displacement to \(Q\): \(R=Q+\overrightarrow{QR}=(-1, 1, 1)+(-6, 8, 0)=(-7, 9, 1)\).

Answer

a) \(\overrightarrow{PQ}=\begin{pmatrix}-3\\4\\0\end{pmatrix}\), \(\lVert\overrightarrow{PQ}\rVert=5\) b) \(R=(-7, 9, 1)\)

52775312

The vectors \(\mathbf{a}=\begin{pmatrix}4\\-2\\4\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}-1\\3\\1\end{pmatrix}\) are given. Find \(\mathbf{c}=2.5\mathbf{a}+6\mathbf{b}\) and \(\lVert\mathbf{c}\rVert\).

Hints

- Multiply every component by its scalar. - Add the scaled vectors componentwise. - Use the magnitude formula after finding \(\mathbf{c}\).

Solution

1. \(2.5\mathbf{a}=\begin{pmatrix}10\\-5\\10\end{pmatrix}\) and \(6\mathbf{b}=\begin{pmatrix}-6\\18\\6\end{pmatrix}\). 2. \(\mathbf{c}=\begin{pmatrix}10\\-5\\10\end{pmatrix}+\begin{pmatrix}-6\\18\\6\end{pmatrix} =\begin{pmatrix}4\\13\\16\end{pmatrix}\). 3. \(\lVert\mathbf{c}\rVert=\sqrt{4^2+13^2+16^2}=\sqrt{441}=21\).

Answer

\(\mathbf{c}=\begin{pmatrix}4\\13\\16\end{pmatrix}\), \(\lVert\mathbf{c}\rVert=21\)

52775612

In rectangular prism \(ABCDEFGH\), let \(\vec{u}=\overrightarrow{AB}\), \(\vec{v}=\overrightarrow{AD}\), and \(\vec{w}=\overrightarrow{AE}\). Point \(M\) is the midpoint of \(\overline{FG}\). Express \(\overrightarrow{DF}\) and \(\overrightarrow{AM}\) in terms of \(\vec{u}\), \(\vec{v}\), and \(\vec{w}\).

Hints

- Decompose each vector into a path along edges. - Opposite parallel edges represent the same vector. - A vector to the midpoint of an edge is one-half of the full edge vector.

Solution

1. Follow the path \(D\to A\to B\to F\): \(\overrightarrow{DF}=\overrightarrow{DA}+\overrightarrow{AB}+\overrightarrow{BF}=-\vec{v}+\vec{u}+\vec{w}\). 2. Therefore, \(\overrightarrow{DF}=\vec{u}-\vec{v}+\vec{w}\). 3. Follow the path \(A\to E\to F\to M\): \(\overrightarrow{AM}=\overrightarrow{AE}+\overrightarrow{EF}+\overrightarrow{FM}\). 4. Since \(M\) is the midpoint of \(\overline{FG}\), \(\overrightarrow{FM}=\frac12\overrightarrow{FG}=\frac12\vec{v}\). 5. Thus, \(\overrightarrow{AM}=\vec{w}+\vec{u}+\frac12\vec{v}=\vec{u}+\frac12\vec{v}+\vec{w}\).

Answer

\(\overrightarrow{DF}=\vec{u}-\vec{v}+\vec{w}\) \(\overrightarrow{AM}=\vec{u}+\frac12\vec{v}+\vec{w}\)

52776012

Find each vector and its magnitude. a) \(\mathbf{c}=\sqrt{3}\begin{pmatrix}\sqrt{3}\\\sqrt{6}\\-3\end{pmatrix}\) b) \(\mathbf{d}=\frac{1}{2}\begin{pmatrix}4\\2\\-6\end{pmatrix}+\begin{pmatrix}1\\2\\9\end{pmatrix}\)

Hints

- Apply the product rules for square roots. - Perform scalar multiplication before vector addition. - Square each component carefully when finding magnitude.

Solution

1. In part a), \(\mathbf{c}=\begin{pmatrix}3\\3\sqrt{2}\\-3\sqrt{3}\end{pmatrix}\). Then \(\lVert\mathbf{c}\rVert=\sqrt{3^2+(3\sqrt{2})^2+(-3\sqrt{3})^2} =\sqrt{54}=3\sqrt{6}\). 2. In part b), \(\mathbf{d}=\begin{pmatrix}2\\1\\-3\end{pmatrix}+\begin{pmatrix}1\\2\\9\end{pmatrix} =\begin{pmatrix}3\\3\\6\end{pmatrix}\). Then \(\lVert\mathbf{d}\rVert=\sqrt{3^2+3^2+6^2}=\sqrt{54}=3\sqrt{6}\).

Answer

a) \(\mathbf{c}=\begin{pmatrix}3\\3\sqrt{2}\\-3\sqrt{3}\end{pmatrix}\), \(\lVert\mathbf{c}\rVert=3\sqrt{6}\) b) \(\mathbf{d}=\begin{pmatrix}3\\3\\6\end{pmatrix}\), \(\lVert\mathbf{d}\rVert=3\sqrt{6}\)

52776412

The parallel vectors \(\mathbf{u}=\begin{pmatrix}a\\-6\\12\end{pmatrix}\) and \(\mathbf{v}=\begin{pmatrix}2\\4\\-8\end{pmatrix}\) satisfy \(\mathbf{u}=s\mathbf{v}\). 1. Find the scale factor \(s\). 2. Use it to find \(a\).

Hints

- Begin with components that contain no unknowns. - Parallel vectors with a specified orientation are scalar multiples. - Use the same scale factor in every component. - Check the factor with another known component.

Solution

1. Compare the second components: \(-6=4s\), so \(s=-1.5\). 2. The third components confirm the result because \(12=(-1.5)\cdot(-8)\). 3. From the first components, \(a=2s=2\cdot(-1.5)=-3\).

Answer

1. \(s=-1.5\) 2. \(a=-3\)

52776712

Simplify each vector expression. a) \(3(\mathbf{x}-2\mathbf{y})+4(\mathbf{y}-\mathbf{z})-2(\mathbf{x}-2\mathbf{z})\) b) \(\frac{1}{2}(4\mathbf{a}-6\mathbf{b})-(\mathbf{a}-5\mathbf{b})+3\mathbf{a}\) c) \(5\mathbf{u}-[2(\mathbf{u}+3\mathbf{v})-4(\mathbf{v}-\mathbf{u})]\)

Hints

- Distribute scalar factors to every term. - A negative sign before parentheses changes every sign inside. - Combine coefficients of the same vector after removing parentheses.

Solution

1. In part a), distribute and combine coefficients: \(3\mathbf{x}-6\mathbf{y}+4\mathbf{y}-4\mathbf{z}-2\mathbf{x}+4\mathbf{z} =\mathbf{x}-2\mathbf{y}\). 2. In part b), \(2\mathbf{a}-3\mathbf{b}-\mathbf{a}+5\mathbf{b}+3\mathbf{a} =4\mathbf{a}+2\mathbf{b}\). 3. In part c), first simplify the brackets: \(2\mathbf{u}+6\mathbf{v}-4\mathbf{v}+4\mathbf{u} =6\mathbf{u}+2\mathbf{v}\). Then \(5\mathbf{u}-(6\mathbf{u}+2\mathbf{v})=-\mathbf{u}-2\mathbf{v}\).

Answer

a) \(\mathbf{x}-2\mathbf{y}\) b) \(4\mathbf{a}+2\mathbf{b}\) c) \(-\mathbf{u}-2\mathbf{v}\)

52776812

Use the distributive property and combine like vector terms. a) \(-(\mathbf{r}-\mathbf{s})+2(\mathbf{r}+\mathbf{s})-3(\mathbf{s}-\mathbf{r})\) b) \(2(3\mathbf{a}-\mathbf{b}+4\mathbf{c})-5(\mathbf{a}+0.4\mathbf{b}-\mathbf{c})\)

Hints

- A negative sign before parentheses reverses all signs inside. - Multiply each vector term by the scalar outside its parentheses. - Terms with coefficient \(0\) disappear from the final expression.

Solution

1. In part a), \(-\mathbf{r}+\mathbf{s}+2\mathbf{r}+2\mathbf{s}-3\mathbf{s}+3\mathbf{r} =4\mathbf{r}\). 2. In part b), \(6\mathbf{a}-2\mathbf{b}+8\mathbf{c}-5\mathbf{a}-2\mathbf{b}+5\mathbf{c} =\mathbf{a}-4\mathbf{b}+13\mathbf{c}\).

Answer

a) \(4\mathbf{r}\) b) \(\mathbf{a}-4\mathbf{b}+13\mathbf{c}\)

52776912

The vector \(\mathbf{a}=\begin{pmatrix}4\\-4\\7\end{pmatrix}\) is multiplied by a real scalar \(k\) to form \(\mathbf{b}=k\mathbf{a}\). Find all values of \(k\) for which \(\lVert\mathbf{b}\rVert=27\).

Hints

- Find the magnitude of \(\mathbf{a}\). - Scalar multiplication changes magnitude by a factor of \(|k|\). - Both a positive and a negative scalar can produce the same magnitude. - Consider how a negative scalar changes direction but not length.

Solution

1. \(\lVert\mathbf{a}\rVert=\sqrt{4^2+(-4)^2+7^2}=\sqrt{81}=9\). 2. Use \(\lVert k\mathbf{a}\rVert=|k|\lVert\mathbf{a}\rVert\): \(9|k|=27\). 3. Thus, \(|k|=3\), so \(k=3\) or \(k=-3\).

Answer

\(k=3\) or \(k=-3\)

52777512

A point \(T\) lies on \(\overline{AB}\) and satisfies \(\overrightarrow{AT}=k\overrightarrow{AB}\), where \(0\leq k\leq1\). a) Show that \(\overrightarrow{OT}=(1-k)\overrightarrow{OA}+k\overrightarrow{OB}\). b) Point \(T\) divides the segment from \(A(2, -1, 4)\) to \(B(6, 7, 0)\) in the ratio \(1:3\). Use the formula from part a to find \(T\).

Hints

- Express the path from \(O\) to \(T\) through \(A\). - Replace \(\overrightarrow{AB}\) with a difference of position vectors. - Convert the \(1:3\) ratio into the fraction of the full segment measured from \(A\).

Solution

1. By vector addition, \(\overrightarrow{OT}=\overrightarrow{OA}+\overrightarrow{AT}\). 2. Substitute \(\overrightarrow{AT}=k\overrightarrow{AB}\) and \(\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}\): \(\overrightarrow{OT}=\overrightarrow{OA}+k(\overrightarrow{OB}-\overrightarrow{OA})\). 3. Distribute and combine terms: \(\overrightarrow{OT}=(1-k)\overrightarrow{OA}+k\overrightarrow{OB}\). 4. A \(1:3\) division from \(A\) gives \(k=\frac14\). 5. Therefore, \(\overrightarrow{OT}=\frac34\begin{pmatrix}2\\-1\\4\end{pmatrix}+\frac14\begin{pmatrix}6\\7\\0\end{pmatrix}=\begin{pmatrix}3\\1\\3\end{pmatrix}\). 6. Thus, \(T(3, 1, 3)\).

Answer

a) \(\overrightarrow{OT}=(1-k)\overrightarrow{OA}+k\overrightarrow{OB}\) b) \(T(3, 1, 3)\)

52777612

In triangle \(ABC\), point \(M_a\) is the midpoint of \(\overline{BC}\). The centroid \(G\) lies on median \(\overline{AM_a}\) and divides it in a \(2:1\) ratio measured from \(A\). a) Express \(\overrightarrow{OM_a}\) in terms of \(\overrightarrow{OB}\) and \(\overrightarrow{OC}\). b) Use part a and the division ratio to derive \(\overrightarrow{OG}=\frac13(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})\).

Hints

- Write the position vector of a midpoint as an average. - A \(2:1\) division means the centroid is two-thirds of the way from the vertex to the midpoint. - Substitute the midpoint expression and combine coefficients.

Solution

1. Since \(M_a\) is the midpoint of \(\overline{BC}\), \(\overrightarrow{OM_a}=\frac12(\overrightarrow{OB}+\overrightarrow{OC})\). 2. The \(2:1\) ratio gives \(\overrightarrow{AG}=\frac23\overrightarrow{AM_a}\). 3. Therefore, \(\overrightarrow{OG}=\overrightarrow{OA}+\frac23(\overrightarrow{OM_a}-\overrightarrow{OA})=\frac13\overrightarrow{OA}+\frac23\overrightarrow{OM_a}\). 4. Substitute the midpoint expression: \(\overrightarrow{OG}=\frac13\overrightarrow{OA}+\frac23\left[\frac12(\overrightarrow{OB}+\overrightarrow{OC})\right]\). 5. Simplifying gives \(\overrightarrow{OG}=\frac13(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})\).

Answer

a) \(\overrightarrow{OM_a}=\frac12(\overrightarrow{OB}+\overrightarrow{OC})\) b) \(\overrightarrow{OG}=\frac13(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})\)

52778212

In triangle \(ABC\), point \(D\) lies on \(\overline{AC}\) and point \(E\) lies on \(\overline{BC}\). Suppose \(\overrightarrow{CD}=\frac14\overrightarrow{CA}\) and \(\overrightarrow{CE}=\frac14\overrightarrow{CB}\). Express \(\overrightarrow{DE}\) in terms of \(\overrightarrow{AB}\), and describe how \(\overline{DE}\) is related to \(\overline{AB}\).

Hints

- Express \(\overrightarrow{DE}\) using a path through \(C\). - Relate \(\overrightarrow{CA}\), \(\overrightarrow{CB}\), and \(\overrightarrow{AB}\). - Interpret a positive scalar multiple geometrically.

Solution

1. Use a path through \(C\): \(\overrightarrow{DE}=\overrightarrow{DC}+\overrightarrow{CE}\). 2. Since \(\overrightarrow{DC}=-\frac14\overrightarrow{CA}\), \(\overrightarrow{DE}=\frac14(\overrightarrow{CB}-\overrightarrow{CA})\). 3. Because \(\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{CB}\), \(\overrightarrow{CB}-\overrightarrow{CA}=\overrightarrow{AB}\). 4. Therefore, \(\overrightarrow{DE}=\frac14\overrightarrow{AB}\). 5. The positive scalar multiple shows that \(\overline{DE}\) is parallel to \(\overline{AB}\), has the same direction, and is one-fourth as long.

Answer

\(\overrightarrow{DE}=\frac14\overrightarrow{AB}\). Thus, \(\overline{DE}\parallel\overline{AB}\) and \(DE=\frac14 AB\).

53028212

Investigate scalar-multiple relationships between vectors. a) Let \(\mathbf{u}=\begin{pmatrix}5\\0\\-2\end{pmatrix}\) and \(\mathbf{v}=\begin{pmatrix}10\\2\\-4\end{pmatrix}\). Show that \(\mathbf{v}\) is not a scalar multiple of \(\mathbf{u}\). b) Find \(m\) and \(n\) so that \(\mathbf{p}=\begin{pmatrix}m\\n\\15\end{pmatrix}\) is a scalar multiple of \(\mathbf{q}=\begin{pmatrix}-2\\4\\5\end{pmatrix}\). c) Evaluate the statement: “If corresponding components of two vectors have the same sign, then one vector is a positive scalar multiple of the other.” Justify your answer.

Hints

- A zero component remains zero under scalar multiplication. - The same scale factor must satisfy all components. - Find the factor in part b) from the third component. - One counterexample disproves a universal statement.

Solution

1. In part a), the first and third components would require a scale factor of \(2\). But the second component would require \(2=2\cdot0\), which is impossible. Therefore, \(\mathbf{v}\) is not a scalar multiple of \(\mathbf{u}\). 2. In part b), \(15=5k\) gives \(k=3\). Thus, \(m=3\cdot(-2)=-6\) and \(n=3\cdot4=12\). 3. The statement is false. For example, \(\mathbf{a}=\begin{pmatrix}1\\2\\3\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}3\\2\\1\end{pmatrix}\) have positive corresponding components, but their component ratios are not equal. Therefore, neither is a scalar multiple of the other.

Answer

a) No scale factor exists because the second components are inconsistent. b) \(m=-6\), \(n=12\) c) False; for example, \(\begin{pmatrix}1\\2\\3\end{pmatrix}\) and \(\begin{pmatrix}3\\2\\1\end{pmatrix}\) are not scalar multiples.

53030212

In quadrilateral \(ABCD\), \(M\) is the midpoint of \(\overline{AB}\) and \(N\) is the midpoint of \(\overline{CD}\). Prove that \(\overrightarrow{MN}=\frac12(\overrightarrow{AD}+\overrightarrow{BC})\).

Hints

- Write two different vector paths from \(M\) to \(N\). - Add the two path equations. - Use the fact that vectors from a midpoint to the two endpoints are opposites.

Solution

1. Use two paths from \(M\) to \(N\): \(\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AD}+\overrightarrow{DN}\) and \(\overrightarrow{MN}=\overrightarrow{MB}+\overrightarrow{BC}+\overrightarrow{CN}\). 2. Add the equations: \(2\overrightarrow{MN}=(\overrightarrow{MA}+\overrightarrow{MB})+\overrightarrow{AD}+\overrightarrow{BC}+(\overrightarrow{DN}+\overrightarrow{CN})\). 3. Since \(M\) and \(N\) are midpoints, \(\overrightarrow{MA}+\overrightarrow{MB}=\vec{0}\) and \(\overrightarrow{DN}+\overrightarrow{CN}=\vec{0}\). 4. Therefore, \(2\overrightarrow{MN}=\overrightarrow{AD}+\overrightarrow{BC}\), so \(\overrightarrow{MN}=\frac12(\overrightarrow{AD}+\overrightarrow{BC})\).

Answer

\(\overrightarrow{MN}=\frac12(\overrightarrow{AD}+\overrightarrow{BC})\)

53031512

Let \(\vec{u},\vec{v}\in\mathbb{R}^3\) and \(r\in\mathbb{R}\). 1. Use components to prove that \(-\bigl(r(\vec{u}+\vec{v})\bigr)=(-r)\vec{u}+(-r)\vec{v}\). 2. Explain geometrically how the negative sign affects the resulting vectors.

Hints

- Write each vector in component form. - Apply the distributive property to each component. - What does multiplying a vector by \(-1\) do to its direction?

Solution

1. Write \(\vec{u}=\begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix}\) and \(\vec{v}=\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}\). 2. The left side is \(-\bigl(r(\vec{u}+\vec{v})\bigr)=\begin{pmatrix}-r(u_1+v_1)\\-r(u_2+v_2)\\-r(u_3+v_3)\end{pmatrix}\). 3. Distributing in each component gives \(\begin{pmatrix}-ru_1-rv_1\\-ru_2-rv_2\\-ru_3-rv_3\end{pmatrix}=\begin{pmatrix}-ru_1\\-ru_2\\-ru_3\end{pmatrix}+\begin{pmatrix}-rv_1\\-rv_2\\-rv_3\end{pmatrix}=(-r)\vec{u}+(-r)\vec{v}\). 4. Multiplication by \(-1\) reverses the direction of each nonzero scaled vector. Thus, \((-r)\vec{u}\) and \((-r)\vec{v}\) point opposite \(r\vec{u}\) and \(r\vec{v}\), respectively.

Answer

1. Componentwise distribution gives \(-r(u_i+v_i)=-ru_i-rv_i\), so \(-\bigl(r(\vec{u}+\vec{v})\bigr)=(-r)\vec{u}+(-r)\vec{v}\). 2. The negative sign reverses the direction of each nonzero scaled vector.

53033812

In a quadrilateral \(ABCD\), points \(E,F,G,H\) are the midpoints of \(\overline{AB},\overline{BC},\overline{CD},\overline{DA}\). 1. Use position vectors to show that \(\overrightarrow{EF}=\overrightarrow{HG}=\frac12\overrightarrow{AC}\). 2. Suppose \(\overrightarrow{AC}=\begin{pmatrix}6\\0\\8\end{pmatrix}\) and \(\overrightarrow{BD}=\begin{pmatrix}0\\6\\8\end{pmatrix}\). Find the side lengths of \(EFGH\). 3. Find the perimeter and classify the parallelogram \(EFGH\).

Hints

- Express each midpoint as an average of endpoint position vectors. - The sides of the midpoint quadrilateral are half the diagonals of the original quadrilateral. - Compare the two diagonal magnitudes.

Solution

1. Since \(\vec{e}=\frac12(\vec{a}+\vec{b})\) and \(\vec{f}=\frac12(\vec{b}+\vec{c})\), \(\overrightarrow{EF}=\vec{f}-\vec{e}=\frac12(\vec{c}-\vec{a})=\frac12\overrightarrow{AC}\). Similarly, \(\overrightarrow{HG}=\frac12\overrightarrow{AC}\). 2. The other pair of sides satisfies \(\overrightarrow{FG}=\overrightarrow{EH}=\frac12\overrightarrow{BD}\). 3. \(\|\overrightarrow{AC}\|=\sqrt{6^2+0^2+8^2}=10\) and \(\|\overrightarrow{BD}\|=\sqrt{0^2+6^2+8^2}=10\). 4. Every side of \(EFGH\) has length \(5\), so its perimeter is \(20\). 5. A parallelogram with four equal sides is a rhombus.

Answer

1. \(\overrightarrow{EF}=\overrightarrow{HG}=\frac12\overrightarrow{AC}\) 2. All four side lengths are \(5\) units. 3. The perimeter is \(20\) units, and \(EFGH\) is a rhombus.

53038612

Solve for \(\vec{x}\): \(3(2\vec{x}-\vec{a})-2(\vec{b}+\vec{x})=5(\vec{a}-\vec{x})+4\vec{b}\).

Hints

- Distribute each scalar first. - Collect coefficients of \(\vec{x}\), \(\vec{a}\), and \(\vec{b}\) separately. - Divide the final vector equation by the coefficient of \(\vec{x}\).

Solution

1. Distribute: \(6\vec{x}-3\vec{a}-2\vec{b}-2\vec{x}=5\vec{a}-5\vec{x}+4\vec{b}\). 2. Combine like terms: \(4\vec{x}-3\vec{a}-2\vec{b}=5\vec{a}-5\vec{x}+4\vec{b}\). 3. Collect the \(\vec{x}\)-terms: \(9\vec{x}=8\vec{a}+6\vec{b}\). 4. Divide by \(9\): \(\vec{x}=\frac89\vec{a}+\frac23\vec{b}\).

Answer

\(\vec{x}=\frac89\vec{a}+\frac23\vec{b}\)

53038912

The points \(P(2, 1, 3)\), \(Q(6, 3, 1)\), and \(R(5, 10, 0)\) are given. a) Find \(S\) so that the diagonals of quadrilateral \(PQRS\) bisect each other. b) Point \(T\) divides \(\overline{PR}\) in a \(2:1\) ratio measured from \(P\). Find \(T\).

Hints

- Use the parallelogram property of bisecting diagonals. - Convert the \(2:1\) ratio into a fraction of the whole segment. - Add that fractional displacement to \(P\).

Solution

1. A quadrilateral whose diagonals bisect each other is a parallelogram, so \(\overrightarrow{PQ}=\overrightarrow{SR}\). 2. \(\overrightarrow{PQ}=Q-P=(4, 2, -2)\), so \(S=R-\overrightarrow{PQ}=(1, 8, 2)\). 3. A \(2:1\) ratio from \(P\) means \(\overrightarrow{PT}=\frac23\overrightarrow{PR}\). 4. \(\overrightarrow{PR}=R-P=(3, 9, -3)\). 5. Therefore, \(T=P+\frac23\overrightarrow{PR}=(2, 1, 3)+(2, 6, -2)=(4, 7, 1)\).

Answer

a) \(S(1, 8, 2)\) b) \(T(4, 7, 1)\)

53039012

Points \(A(1, 2, 4)\), \(B(5, 4, 0)\), and \(C(3, 9, 1)\) are consecutive vertices of trapezoid \(ABCD\). a) Side \(\overline{CD}\) is parallel to \(\overline{AB}\) and half as long. Find \(D\). b) Let \(M\) be the midpoint of \(\overline{AB}\). Determine whether \(Z(3, 6, 1.5)\) lies on \(\overline{MC}\).

Hints

- Translate the parallel-length condition into a vector equation. - Use the midpoint formula for \(M\). - A point lies on a segment when its displacement from one endpoint is a scalar multiple between \(0\) and \(1\) of the full displacement.

Solution

1. With the given vertex order, \(\overrightarrow{DC}=\frac12\overrightarrow{AB}\). 2. \(\overrightarrow{AB}=B-A=(4, 2, -4)\), so \(\overrightarrow{DC}=(2, 1, -2)\). 3. Thus, \(D=C-\overrightarrow{DC}=(1, 8, 3)\). 4. The midpoint of \(\overline{AB}\) is \(M(3, 3, 2)\). 5. \(\overrightarrow{MZ}=(0, 3, -0.5)\) and \(\overrightarrow{MC}=(0, 6, -1)\). 6. Since \(\overrightarrow{MZ}=0.5\overrightarrow{MC}\) and \(0.5\) is between \(0\) and \(1\), \(Z\) lies on \(\overline{MC}\).

Answer

a) \(D(1, 8, 3)\) b) Yes. Since \(\overrightarrow{MZ}=0.5\overrightarrow{MC}\), \(Z\) lies on \(\overline{MC}\).

53039312

In triangle \(ABC\), \(A(2, 5, -1)\), \(B(-4, 1, 3)\), and the centroid is \(G(1, 2, 2)\). a) Find \(C\). b) Find the midpoint \(M\) of \(\overline{AB}\). c) Verify that \(G\) divides median \(\overline{CM}\) in a \(2:1\) ratio measured from \(C\).

Hints

- Rearrange the centroid formula to solve for the missing vertex. - Use the midpoint formula for \(\overline{AB}\). - Compare the two displacement vectors along the median.

Solution

1. The centroid formula gives \(G=\frac13(A+B+C)\), so \(C=3G-A-B\). 2. \(C=3(1, 2, 2)-(2, 5, -1)-(-4, 1, 3)=(5, 0, 4)\). 3. The midpoint of \(\overline{AB}\) is \(M=\left(\frac{2+(-4)}2,\frac{5+1}2,\frac{-1+3}2\right)=(-1, 3, 1)\). 4. \(\overrightarrow{CG}=G-C=(-4, 2, -2)\), while \(\overrightarrow{GM}=M-G=(-2, 1, -1)\). 5. Since \(\overrightarrow{CG}=2\overrightarrow{GM}\), the ratio is \(CG:GM=2:1\).

Answer

a) \(C(5, 0, 4)\) b) \(M(-1, 3, 1)\) c) \(\overrightarrow{CG}=2\overrightarrow{GM}\), so \(G\) divides \(\overline{CM}\) in a \(2:1\) ratio.

53039412

Triangle \(ABC\) has vertices \(A(3, 1, 4)\), \(B(7, 5, 2)\), and \(C(1, 9, 6)\). Let \(M\) be the midpoint of \(\overline{AB}\) and \(N\) the midpoint of \(\overline{AC}\). a) Find \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\). b) Find \(M\) and \(N\). c) Find \(\overrightarrow{MN}\) and compare it with \(\overrightarrow{BC}\). d) Prove generally, without coordinates, that \(\overrightarrow{MN}=\frac12\overrightarrow{BC}\).

Hints

- Subtract endpoint coordinates to find side vectors. - Use the midpoint formula twice. - For the proof, write both midpoint position vectors and subtract.

Solution

1. \(\overrightarrow{AB}=B-A=(4, 4, -2)\) and \(\overrightarrow{AC}=C-A=(-2, 8, 2)\). 2. \(M=\frac12(A+B)=(5, 3, 3)\) and \(N=\frac12(A+C)=(2, 5, 5)\). 3. \(\overrightarrow{MN}=N-M=(-3, 2, 2)\), while \(\overrightarrow{BC}=C-B=(-6, 4, 4)\). Thus, \(\overrightarrow{MN}=\frac12\overrightarrow{BC}\). 4. In general, if \(\vec{m}=\frac12(\vec{a}+\vec{b})\) and \(\vec{n}=\frac12(\vec{a}+\vec{c})\), then \(\overrightarrow{MN}=\vec{n}-\vec{m}=\frac12(\vec{c}-\vec{b})=\frac12\overrightarrow{BC}\).

Answer

a) \(\overrightarrow{AB}=\begin{pmatrix}4\\4\\-2\end{pmatrix}\), \(\overrightarrow{AC}=\begin{pmatrix}-2\\8\\2\end{pmatrix}\) b) \(M(5, 3, 3)\), \(N(2, 5, 5)\) c) \(\overrightarrow{MN}=\begin{pmatrix}-3\\2\\2\end{pmatrix}=\frac12\overrightarrow{BC}\) d) The midpoint formulas simplify to \(\overrightarrow{MN}=\frac12(\vec{c}-\vec{b})=\frac12\overrightarrow{BC}\).

53039812

A quadrilateral \(ABCD\) has position vectors \(\vec{a},\vec{b},\vec{c},\vec{d}\). 1. Show that if \(\vec{a}+\vec{c}=\vec{b}+\vec{d}\), then \(\overrightarrow{AB}=\overrightarrow{DC}\). 2. Use midpoint vectors to show that diagonals \(\overline{AC}\) and \(\overline{BD}\) have the same midpoint. 3. State the resulting theorem about a quadrilateral's diagonals and its classification.

Hints

- Express each side vector as terminal position vector minus initial position vector. - Compare the midpoint formulas for the two diagonals. - Recall the converse of the parallelogram diagonal property.

Solution

1. From \(\vec{a}+\vec{c}=\vec{b}+\vec{d}\), subtract \(\vec{a}+\vec{d}\) to obtain \(\vec{b}-\vec{a}=\vec{c}-\vec{d}\). Thus, \(\overrightarrow{AB}=\overrightarrow{DC}\). 2. The midpoint vectors are \(\vec{m}_{AC}=\frac12(\vec{a}+\vec{c})\) and \(\vec{m}_{BD}=\frac12(\vec{b}+\vec{d})\). The given equation makes these vectors equal, so the diagonals have the same midpoint and bisect each other. 3. A quadrilateral is a parallelogram if and only if its diagonals bisect each other.

Answer

1. \(\overrightarrow{AB}=\overrightarrow{DC}\) 2. \(\vec{m}_{AC}=\vec{m}_{BD}\) 3. A quadrilateral is a parallelogram if and only if its diagonals bisect each other.

53040612

A corner-cube retroreflector can be modeled by three mutually perpendicular reflecting surfaces: the \(xy\)-plane, the \(xz\)-plane, and the \(yz\)-plane. A light ray with direction vector \(\vec{v}=\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}\) reflects from these planes in that order. a) Write the direction vector after each reflection. b) Show that the final vector is \(k\vec{v}\), and find \(k\). c) Explain how this result describes the operation of a bicycle reflector.

Hints

- At each step, change the sign of the component perpendicular to the reflecting plane. - Compare the final components with the original components. - What does multiplying a vector by \(-1\) do to its direction?

Solution

1. After reflection across the \(xy\)-plane, \(\vec{v}_1=\begin{pmatrix}v_1\\v_2\\-v_3\end{pmatrix}\). 2. After reflection across the \(xz\)-plane, \(\vec{v}_2=\begin{pmatrix}v_1\\-v_2\\-v_3\end{pmatrix}\). 3. After reflection across the \(yz\)-plane, \(\vec{v}_{\text{final}}=\begin{pmatrix}-v_1\\-v_2\\-v_3\end{pmatrix}\). 4. Thus, \(\vec{v}_{\text{final}}=-\vec{v}\), so \(k=-1\). 5. The final ray is parallel to the incoming ray but points in the opposite direction, sending light back toward its source.

Answer

a) \(\vec{v}_1=\begin{pmatrix}v_1\\v_2\\-v_3\end{pmatrix}\), \(\vec{v}_2=\begin{pmatrix}v_1\\-v_2\\-v_3\end{pmatrix}\), and \(\vec{v}_{\text{final}}=\begin{pmatrix}-v_1\\-v_2\\-v_3\end{pmatrix}\) b) \(\vec{v}_{\text{final}}=-\vec{v}\), so \(k=-1\). c) The ray is returned toward the light source.

53040812

Let \(\vec{v}=\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}\). a) Find the image \(\vec{v}'\) under a point reflection across the origin. b) Show, by listing the intermediate vectors, that reflecting \(\vec{v}\) successively across the \(xy\)-, \(xz\)-, and \(yz\)-planes produces the same result.

Hints

- A point reflection across the origin changes every coordinate's sign. - For each coordinate plane, identify the perpendicular coordinate. - Apply the reflections one at a time.

Solution

1. A point reflection across the origin multiplies every component by \(-1\), so \(\vec{v}'=\begin{pmatrix}-v_1\\-v_2\\-v_3\end{pmatrix}\). 2. Reflection across the \(xy\)-plane gives \(\begin{pmatrix}v_1\\v_2\\-v_3\end{pmatrix}\). 3. Reflection across the \(xz\)-plane then gives \(\begin{pmatrix}v_1\\-v_2\\-v_3\end{pmatrix}\). 4. Reflection across the \(yz\)-plane finally gives \(\begin{pmatrix}-v_1\\-v_2\\-v_3\end{pmatrix}\), which equals \(\vec{v}'\).

Answer

a) \(\vec{v}'=\begin{pmatrix}-v_1\\-v_2\\-v_3\end{pmatrix}\) b) The successive vectors are \(\begin{pmatrix}v_1\\v_2\\-v_3\end{pmatrix}\), \(\begin{pmatrix}v_1\\-v_2\\-v_3\end{pmatrix}\), and \(\begin{pmatrix}-v_1\\-v_2\\-v_3\end{pmatrix}\), so the final result equals \(\vec{v}'\).

53051912

The point set \(S\) is given by \(\vec{x}=\begin{pmatrix}4\\0\\-2\end{pmatrix}+\lambda\begin{pmatrix}3\\-2\\1\end{pmatrix}+\mu\begin{pmatrix}m\\4\\-2\end{pmatrix}\), where \(\lambda,\mu,m\in\mathbb{R}\). a) Find \(m\) so that \(S\) describes a line rather than a plane. b) For this value of \(m\), write a simplified parametric equation of the line. c) Determine algebraically whether \(P(10, -4, 0)\) lies on the line.

Hints

- When do two direction vectors generate only one direction? - Compare corresponding components to find the scale factor. - Replace the two dependent parameters with one new parameter. - Test the point in all three coordinate equations.

Solution

1. The point set describes a line when its two direction vectors are linearly dependent. Require \(\begin{pmatrix}m\\4\\-2\end{pmatrix}=k\begin{pmatrix}3\\-2\\1\end{pmatrix}\). 2. The second and third coordinates give \(k=-2\). Therefore, \(m=3\cdot(-2)=-6\). 3. For \(m=-6\), the second direction vector is \(-2\) times the first, so the two parameters combine into one. A simplified equation is \(g:\vec{x}=\begin{pmatrix}4\\0\\-2\end{pmatrix}+t\begin{pmatrix}3\\-2\\1\end{pmatrix}\). 4. For \(P\), the coordinate equations are \(10=4+3t\), \(-4=-2t\), and \(0=-2+t\). Each gives \(t=2\), so \(P\) lies on the line.

Answer

a) \(m=-6\) b) \(g:\vec{x}=\begin{pmatrix}4\\0\\-2\end{pmatrix}+t\begin{pmatrix}3\\-2\\1\end{pmatrix}\) c) Yes. Point \(P\) lies on the line when \(t=2\).

53053512

Points \(A(1, 1, 1)\), \(B(3, 2, 0)\), \(C(0, 4, 2)\), and \(D(2, 5, k)\) are given, where \(k\in\mathbb{R}\). a) Find the value of \(k\) for which line \(AB\) is parallel to line \(CD\). b) Explain why the four points are coplanar for that value of \(k\).

Hints

- Form a direction vector for each line. - Parallel vectors are scalar multiples. - Recall the plane determined by two distinct parallel lines.

Solution

1. a) The direction vectors are \(\overrightarrow{AB}=\langle 2,1,-1\rangle\) and \(\overrightarrow{CD}=\langle 2,1,k-2\rangle\). 2. Parallel lines have direction vectors that are scalar multiples. Since the first components are both \(2\), the scalar is \(1\). 3. Therefore, the third components must also be equal: \(k-2=-1\), so \(k=1\). 4. b) For \(k=1\), the direction vectors are equal. The lines are distinct because point \(C\) is not on line \(AB\). 5. Two distinct parallel lines lie in exactly one plane. Since \(A\) and \(B\) lie on one line and \(C\) and \(D\) lie on the other, all four points are coplanar.

Answer

a) \(k=1\) b) The two lines are distinct and parallel, so they lie in one plane; therefore, \(A\), \(B\), \(C\), and \(D\) are coplanar.

53061412

Planes \(E\) and \(F_{t,s}\) are given by \(E:3x+2y-4z=6\), \(F_{t,s}:9x+6y+tz=s\), where \(t,s\in\mathbb{R}\). a) Find \(t\) so that the planes are parallel or identical. b) For that value of \(t\), find \(s\) so that the planes are identical. c) Describe the relationship between the planes when \(t=-12\) and \(s=10\).

Hints

- Compare the normal vectors as scalar multiples. - Identical equations require the same scalar factor on both sides. - Parallel normals with nonproportional constants give strictly parallel planes.

Solution

1. a) The normal vectors are \(\vec{n}_E=\langle 3,2,-4\rangle\) and \(\vec{n}_F=\langle 9,6,t\rangle\). 2. Parallel planes have normal vectors that are scalar multiples. The first two components show that \(\vec{n}_F=3\vec{n}_E\). 3. Therefore, \(t=3(-4)=-12\). 4. b) For the equations to describe the same plane, the right side must have the same scalar factor. Thus \(s=3(6)=18\). 5. c) When \(t=-12\), the normals are parallel. When \(s=10\), the right side is not \(18\), so the equations do not describe the same plane. 6. Therefore, the planes are strictly parallel.

Answer

a) \(t=-12\) b) \(s=18\) c) The planes are strictly parallel.

53062112

Planes \(E_1\) and \(E_2\) are given by \(E_1: 5x-y+2z=10\), \(E_2: -10x+2y-4z=5\). a) Determine their relationship. b) Plane \(E_3\) is parallel to \(E_1\) and contains \(P(2, -3, 1)\). Find an equation of \(E_3\). c) Write the family of all planes parallel to \(E_1\).

Hints

- Compare the normal vectors. - Apply the same scale factor to the entire equation to test whether the planes are identical. - A parallel plane can use the same normal vector. - Substitute the given point to find the constant.

Solution

1. a) The normal vectors satisfy \(\langle -10,2,-4\rangle=-2\langle 5,-1,2\rangle\), so the planes are parallel or identical. 2. Multiplying the equation of \(E_1\) by \(-2\) gives \(-10x+2y-4z=-20\), not \(5\). Therefore the planes are strictly parallel. 3. b) A plane parallel to \(E_1\) can use the same normal vector, so write \(5x-y+2z=d\). 4. Substitute \(P(2, -3, 1)\): \(d=5(2)-(-3)+2(1)=15\). Thus \(E_3:5x-y+2z=15\). 5. c) Every plane parallel to \(E_1\) has the form \(5x-y+2z=d\), where \(d\in\mathbb{R}\).

Answer

a) \(E_1\) and \(E_2\) are strictly parallel. b) \(E_3:5x-y+2z=15\) c) \(5x-y+2z=d\), where \(d\in\mathbb{R}\)

53062712

Planes \(E_1\) and \(E_2\) are given by \(E_1:6x-4y+2z=12\), \(E_2:ax+6y+bz=c\). Find all real values of \(a\), \(b\), and \(c\) for which the planes are a) identical, b) strictly parallel, c) intersecting in a line.

Hints

- Compare the normal vectors as scalar multiples. - Use the fixed y-coefficient to find the scalar factor. - Identical equations require the constant to have the same factor. - Nonparallel planes intersect in a line.

Solution

1. The normal vectors are \(\vec{n}_1=\langle 6,-4,2\rangle\) and \(\vec{n}_2=\langle a,6,b\rangle\). 2. For the planes to be parallel or identical, the normals must be scalar multiples. Since the y-component satisfies \(6=q(-4)\), the scalar is \(q=-\frac32\). 3. Therefore, \(a=-\frac32(6)=-9\) and \(b=-\frac32(2)=-3\). 4. a) The planes are identical when the constant has the same scalar factor: \(c=-\frac32(12)=-18\). 5. b) The planes are strictly parallel when \(a=-9\), \(b=-3\), and \(c\neq-18\). 6. c) If \(a\neq-9\) or \(b\neq-3\), the normals are not scalar multiples. Two nonparallel planes in space intersect in a line, regardless of \(c\).

Answer

a) \(a=-9\), \(b=-3\), \(c=-18\) b) \(a=-9\), \(b=-3\), \(c\neq-18\) c) \(a\neq-9\) or \(b\neq-3\), with any real \(c\)

52778312

In tetrahedron \(ABCD\), point \(G\) is the centroid of triangle \(ABC\). Point \(S\) lies on \(\overline{DG}\) and divides it in a \(3:1\) ratio measured from \(D\). 1. Express \(\overrightarrow{OS}\) in terms of the position vectors of \(A\), \(B\), \(C\), and \(D\). 2. Let \(M_{AB}\) and \(M_{CD}\) be the midpoints of \(\overline{AB}\) and \(\overline{CD}\). Prove that \(S\) is the midpoint of \(\overline{M_{AB}M_{CD}}\).

Hints

- Begin with the centroid formula for triangle \(ABC\). - Convert the \(3:1\) ratio into a fraction of \(\overline{DG}\). - Find each edge midpoint, then average their position vectors.

Solution

1. The centroid of triangle \(ABC\) has position vector \(\overrightarrow{OG}=\frac13(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})\). 2. Since \(S\) is three-fourths of the way from \(D\) to \(G\), \(\overrightarrow{OS}=\overrightarrow{OD}+\frac34(\overrightarrow{OG}-\overrightarrow{OD})\). 3. Substitution and simplification give \(\overrightarrow{OS}=\frac14(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD})\). 4. The edge midpoints satisfy \(\overrightarrow{OM_{AB}}=\frac12(\overrightarrow{OA}+\overrightarrow{OB})\) and \(\overrightarrow{OM_{CD}}=\frac12(\overrightarrow{OC}+\overrightarrow{OD})\). 5. The midpoint of \(\overline{M_{AB}M_{CD}}\) has position vector \(\frac12(\overrightarrow{OM_{AB}}+\overrightarrow{OM_{CD}})=\frac14(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD})\). 6. This equals \(\overrightarrow{OS}\), so \(S\) is that midpoint.

Answer

1. \(\overrightarrow{OS}=\frac14(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD})\) 2. The midpoint of \(\overline{M_{AB}M_{CD}}\) has the same position vector, so it is \(S\).

52778412

A point \(P\) associated with triangle \(ABC\) is defined by \(3\overrightarrow{PA}+2\overrightarrow{PB}+\overrightarrow{PC}=\vec{0}\). 1. Express \(\overrightarrow{OP}\) in terms of \(\overrightarrow{OA}\), \(\overrightarrow{OB}\), and \(\overrightarrow{OC}\). 2. Show that \(P\) lies on a segment \(\overline{AM}\), where \(M\) lies on \(\overline{BC}\). State the ratios \(BM:MC\) and \(AP:PM\).

Hints

- Rewrite each vector from \(P\) as a difference of position vectors. - Group the coefficients of \(B\) and \(C\) so they define a point on \(\overline{BC}\). - A weighted average with coefficients \(\frac12\) and \(\frac12\) is a midpoint.

Solution

1. Replace each displacement vector by a difference of position vectors: \(3(\overrightarrow{OA}-\overrightarrow{OP})+2(\overrightarrow{OB}-\overrightarrow{OP})+(\overrightarrow{OC}-\overrightarrow{OP})=\vec{0}\). 2. Combining terms gives \(6\overrightarrow{OP}=3\overrightarrow{OA}+2\overrightarrow{OB}+\overrightarrow{OC}\). 3. Therefore, \(\overrightarrow{OP}=\frac12\overrightarrow{OA}+\frac13\overrightarrow{OB}+\frac16\overrightarrow{OC}\). 4. Define \(M\) by \(\overrightarrow{OM}=\frac23\overrightarrow{OB}+\frac13\overrightarrow{OC}\). Then \(M\) lies on \(\overline{BC}\) and \(BM:MC=1:2\). 5. Now \(\overrightarrow{OP}=\frac12\overrightarrow{OA}+\frac12\overrightarrow{OM}\). Thus, \(P\) is the midpoint of \(\overline{AM}\), so \(AP:PM=1:1\).

Answer

1. \(\overrightarrow{OP}=\frac12\overrightarrow{OA}+\frac13\overrightarrow{OB}+\frac16\overrightarrow{OC}\) 2. \(BM:MC=1:2\) and \(AP:PM=1:1\).

53033712

A quadrilateral \(ABCD\) in three-dimensional space has position vectors \(\vec{a},\vec{b},\vec{c},\vec{d}\). Let \(P,Q,R,S\) be the midpoints of \(\overline{AB},\overline{BC},\overline{CD},\overline{DA}\), respectively. 1. Show that \(\overrightarrow{PQ}=\frac12(\vec{c}-\vec{a})\). 2. Prove that \(PQRS\) is a parallelogram by showing \(\overrightarrow{PQ}=\overrightarrow{SR}\). 3. Explain why \(PQRS\) is planar even when \(ABCD\) is not. 4. Describe what happens when \(A=C\).

Hints

- Write each midpoint as the average of its endpoints. - Compare the vectors of opposite sides. - A parallelogram generated by two vectors lies in one plane. - Examine what happens when a side vector becomes the zero vector.

Solution

1. The midpoint vectors are \(\vec{p}=\frac12(\vec{a}+\vec{b})\) and \(\vec{q}=\frac12(\vec{b}+\vec{c})\). Thus, \(\overrightarrow{PQ}=\vec{q}-\vec{p}=\frac12(\vec{c}-\vec{a})\). 2. Similarly, \(\vec{s}=\frac12(\vec{a}+\vec{d})\) and \(\vec{r}=\frac12(\vec{c}+\vec{d})\), so \(\overrightarrow{SR}=\vec{r}-\vec{s}=\frac12(\vec{c}-\vec{a})=\overrightarrow{PQ}\). Therefore, one pair of opposite sides is equal and parallel, so \(PQRS\) is a parallelogram. 3. Points \(P,Q,S\) determine a plane unless the figure is degenerate. Since \(R=S+\overrightarrow{PQ}\), point \(R\) lies in that same plane. 4. If \(A=C\), then \(\overrightarrow{PQ}=\vec{0}\), so \(P=Q\) and \(S=R\). The parallelogram degenerates to at most a segment, or to a point if the remaining endpoints also coincide.

Answer

1. \(\overrightarrow{PQ}=\frac12(\vec{c}-\vec{a})\) 2. \(\overrightarrow{SR}=\frac12(\vec{c}-\vec{a})=\overrightarrow{PQ}\), so \(PQRS\) is a parallelogram. 3. The fourth vertex is obtained from the other three by a parallelogram translation, so all four are coplanar. 4. If \(A=C\), then \(P=Q\) and \(S=R\); the figure degenerates to at most a segment.

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