Free math worksheets

- Scalar multiplication changes every component by the same factor. - Apply the scalar to the first component. - Apply the same scalar to the second component.

Multiply each component by \(3\): \(3\mathbf{v}=\begin{pmatrix}6\\-3\end{pmatrix}\).

\(\begin{pmatrix}6\\-3\end{pmatrix}\)

- Compare the endpoint coordinates of the two vectors. - A scalar multiple changes both components by the same factor. - The opposite directions tell you something about the sign of the scalar.

From the diagram, \(\mathbf{u}=\begin{pmatrix}2\\1\end{pmatrix}\) and \(\mathbf{w}=\begin{pmatrix}-4\\-2\end{pmatrix}\). Both components of \(\mathbf{w}\) are \(-2\) times the corresponding components of \(\mathbf{u}\), so \(k=-2\).

\(k=-2\)

- 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 \(3\mathbf{v}\).

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 \(Q\): \(R=Q+2\mathbf{v}=(1, 4, 6)+(-6, 12, 2)=(-5, 16, 8)\).

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

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

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

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

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

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, \(\lVert\mathbf{b}\rVert=\sqrt{4^2+(-3)^2}=5\).

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

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

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

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

- 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.

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).

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

- 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.

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, \(a=2s=2\cdot(-1.5)=-3\).

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

- Compute each coordinate separately. - Watch the sign when subtracting a negative component. - Use the three-dimensional magnitude formula after finding the vector.

1. Combine corresponding components: \(\vec{v}=\begin{pmatrix}2\cdot1-3\cdot1+3\\2\cdot0-3\cdot(-1)+0\\2\cdot4-3\cdot2+4\end{pmatrix}=\begin{pmatrix}2\\3\\6\end{pmatrix}\). 2. Find the magnitude: \(\|\vec{v}\|=\sqrt{2^2+3^2+6^2}=\sqrt{49}=7\).

\(\vec{v}=\begin{pmatrix}2\\3\\6\end{pmatrix}\); \(\|\vec{v}\|=7\)

- Perform the scalar multiplications first. - Apply the subtraction to every component of the last vector. - Use the magnitude formula after simplifying the vector.

1. Scale each vector: \(3\begin{pmatrix}-1\\2\\1\end{pmatrix}=\begin{pmatrix}-3\\6\\3\end{pmatrix}\), \(2\begin{pmatrix}3\\0\\4\end{pmatrix}=\begin{pmatrix}6\\0\\8\end{pmatrix}\), and \(4\begin{pmatrix}0.5\\0.5\\0.75\end{pmatrix}=\begin{pmatrix}2\\2\\3\end{pmatrix}\). 2. Combine the results: \(\vec{w}=\begin{pmatrix}-3+6-2\\6+0-2\\3+8-3\end{pmatrix}=\begin{pmatrix}1\\4\\8\end{pmatrix}\). 3. Find the magnitude: \(\|\vec{w}\|=\sqrt{1^2+4^2+8^2}=\sqrt{81}=9\).

\(\vec{w}=\begin{pmatrix}1\\4\\8\end{pmatrix}\); \(\|\vec{w}\|=9\)

- Evaluate the expression inside the brackets first. - Multiply every component by \(0.5\). - Use the magnitude formula for a vector in three dimensions.

1. Add inside the brackets: \(\begin{pmatrix}1\\-5\\12\end{pmatrix}+\begin{pmatrix}5\\13\\12\end{pmatrix}=\begin{pmatrix}6\\8\\24\end{pmatrix}\). 2. Multiply by \(0.5\): \(\vec{a}=\begin{pmatrix}3\\4\\12\end{pmatrix}\). 3. Find the magnitude: \(\|\vec{a}\|=\sqrt{3^2+4^2+12^2}=\sqrt{169}=13\).

\(\vec{a}=\begin{pmatrix}3\\4\\12\end{pmatrix}\); \(\|\vec{a}\|=13\)

- Notice that both terms contain the same vector. - Combine the scalar coefficients before multiplying. - Squaring removes the sign of a negative component in the magnitude formula.

1. Combine the scalar coefficients: \(\vec{b}=(1.4+0.6)\begin{pmatrix}2\\5\\-14\end{pmatrix}=2\begin{pmatrix}2\\5\\-14\end{pmatrix}\). 2. Thus, \(\vec{b}=\begin{pmatrix}4\\10\\-28\end{pmatrix}\). 3. Its magnitude is \(\|\vec{b}\|=\sqrt{4^2+10^2+(-28)^2}=\sqrt{900}=30\).

\(\vec{b}=\begin{pmatrix}4\\10\\-28\end{pmatrix}\); \(\|\vec{b}\|=30\)

- 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.

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 \(z=2\cdot(-4)=-8\).

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

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

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, \(-\mathbf{v}=\begin{pmatrix}-v_1\\-v_2\\-v_3\end{pmatrix}\).

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

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

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 \(C'=(-2, 2, 5)+(2, -4, 0)=(0, -2, 5)\).

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

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

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

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

- 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.

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 \(g:\vec{x}=t\begin{pmatrix}3\\-2\\4\end{pmatrix}\).

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 \(t\in\mathbb{R}\)

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

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.

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

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

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.

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.

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

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.

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

- 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.

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.

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.

- 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.

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.

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

- 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.

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 \(d\in\mathbb{R}\).

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

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

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.

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.

- 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.

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.

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

- 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 \(Q\).

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 \(Q\): \(R=Q+\overrightarrow{QR}=(-1, 1, 1)+(-6, 8, 0)=(-7, 9, 1)\).

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

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

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. \(\lVert\mathbf{c}\rVert=\sqrt{4^2+13^2+16^2}=\sqrt{441}=21\).

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

- 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.

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, \(\overrightarrow{AM}=\vec{w}+\vec{u}+\frac12\vec{v}=\vec{u}+\frac12\vec{v}+\vec{w}\).

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

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

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 \(\lVert\mathbf{d}\rVert=\sqrt{3^2+3^2+6^2}=\sqrt{54}=3\sqrt{6}\).

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

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

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 \(5\mathbf{u}-(6\mathbf{u}+2\mathbf{v})=-\mathbf{u}-2\mathbf{v}\).

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

- 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.

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), \(6\mathbf{a}-2\mathbf{b}+8\mathbf{c}-5\mathbf{a}-2\mathbf{b}+5\mathbf{c} =\mathbf{a}-4\mathbf{b}+13\mathbf{c}\).

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

- 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.

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 \(k=-3\).

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

- 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 \(A\).

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, \(T(3, 1, 3)\).

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

- 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.

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 \(\overrightarrow{OG}=\frac13(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})\).

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

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

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.

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

- 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.

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.

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.

- 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.

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

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

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

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.

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.

- 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.

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.

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.

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

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 \(9\): \(\vec{x}=\frac89\vec{a}+\frac23\vec{b}\).

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

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

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, \(T=P+\frac23\overrightarrow{PR}=(2, 1, 3)+(2, 6, -2)=(4, 7, 1)\).

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

- 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.

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 \(\overline{MC}\).

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

- 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.

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 \(CG:GM=2:1\).

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.

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

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 \(\overrightarrow{MN}=\vec{n}-\vec{m}=\frac12(\vec{c}-\vec{b})=\frac12\overrightarrow{BC}\).

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 \(\overrightarrow{MN}=\frac12(\vec{c}-\vec{b})=\frac12\overrightarrow{BC}\).

- 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.

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.

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.

- 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?

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.

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.

- 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.

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 \(\vec{v}'\).

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 \(\vec{v}'\).

- 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.

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.

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 \(t=2\).

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

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.

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.

- 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.

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.

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

- 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.

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 \(d\in\mathbb{R}\).

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

- 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.

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 \(c\).

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 \(c\)

- 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.

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.

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 \(S\).

- 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.

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 \(AP:PM=1:1\).

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

- 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.

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.

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.