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

1. Multiply each component by \(3\): \(3\mathbf{v}=\langle6,-3\rangle\).

\(\langle6,-3\rangle\)

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

1. The graph gives \(\mathbf{u}=\langle2,1\rangle\) and \(\mathbf{w}=\langle-4,-2\rangle\). 2. Both components of \(\mathbf{w}\) are \(-2\) times the corresponding components of \(\mathbf{u}\), so \(k=-2\).

\(k=-2\)

- Compare corresponding components. - The same scale factor must work for every component.

1. From the graph, \(\mathbf{u}=\langle2,1\rangle\) and \(\mathbf{w}=\langle6,3\rangle\). 2. Since \(\langle6,3\rangle=3\langle2,1\rangle\), the vectors are scalar multiples. 3. Therefore, \(k=3\).

Yes. \(\mathbf{w}=3\mathbf{u}\), so \(k=3\).

- Scalar multiplication multiplies every component by the same number. - A negative scalar also reverses direction.

1. Multiply every component of \(\mathbf{u}\) by \(-2\): \(-2\mathbf{u}=\langle-4,2,-6\rangle.\)

\(\langle-4,2,-6\rangle\).

- A negative scalar reverses the direction of the vector. - How does the absolute value of a scalar affect the vector's length?

1. Vector \(\mathbf{u}\) runs from the origin to \((2,1)\). 2. Multiplying by \(-2\) reverses its direction and doubles its length, giving components \((-4,-2)\). 3. That is the vector labeled \(\mathbf{v}\), not \(\mathbf{w}\).

\(\mathbf{v}\). It points opposite to \(\mathbf{u}\) and is twice as long.

- Scalar multiplication affects every component. - Does a vector of zero length point in any particular direction?

1. Multiply both components by \(0\): \(0\mathbf{u}=\langle0,0\rangle\). 2. The zero vector has length \(0\) and no defined direction, so it does not have a direction to compare with \(\mathbf{u}\).

\(\langle0,0\rangle\). No: the zero vector has no defined direction.

- Form the displacement vector from \(A\) to \(B\) first. - Collinearity here is equivalent to one displacement vector being a scalar multiple of another. - A single scalar must work in every component.

1. \(\mathbf{v}=\overrightarrow{AB}=\langle2,-4,4\rangle\). 2. \(\overrightarrow{AC}=\langle-1,2,-2\rangle=-\frac12\mathbf{v}\), so \(C\) is collinear with \(A\) and \(B\). 3. \(\overrightarrow{AD}=\langle3,-6,7\rangle\). The first two components would require the scalar \(\frac32\), but \(\frac32\cdot4=6\ne7\). Therefore, \(\overrightarrow{AD}\) is not a scalar multiple of \(\mathbf{v}\), so \(D\) is not collinear with \(A\) and \(B\).

a) Yes; \(\overrightarrow{AC}=-\frac12\mathbf{v}\). b) No; \(\overrightarrow{AD}\) is not a scalar multiple of \(\mathbf{v}\).

- Compute the reference displacement vector once and compare every component with it. - In part a, two known components determine the only possible scalar before the missing coordinate is found. - Opposite directions correspond to a negative scalar.

1. \(\overrightarrow{PQ}=\langle3,3,-3\rangle\). 2. \(\overrightarrow{PR}=\langle a-2,6,-6\rangle\). The second and third components show that the scalar must be \(2\). Therefore, \(a-2=2\cdot3=6\), so \(a=8\). 3. \(\overrightarrow{PS}=\langle-3,-3,3\rangle=-\overrightarrow{PQ}\). Thus, \(\overrightarrow{PS}\) is a scalar multiple of \(\overrightarrow{PQ}\).

a) \(a=8\), with scalar \(2\) b) Yes; \(\overrightarrow{PS}=-\overrightarrow{PQ}\).

- Compute each displacement vector before comparing directions. - Dependence of two nonzero vectors means that one is a scalar multiple of the other. - A single mismatched zero/nonzero component can rule out a scalar multiple immediately.

1. In part a, \(\overrightarrow{AB}=\langle1,2,1\rangle\) and \(\overrightarrow{AC}=\langle3,6,3\rangle=3\overrightarrow{AB}\). The directions are dependent. 2. In part b, \(\overrightarrow{GP}=\langle-1,1,0\rangle\). This cannot be a scalar multiple of \(\langle0,1,1\rangle\) because its first component is nonzero while every scalar multiple of \(\mathbf{u}\) has first component zero. The directions are independent.

a) Dependent; \(\overrightarrow{AC}=3\overrightarrow{AB}\). b) Independent; no scalar multiple of \(\langle0,1,1\rangle\) equals \(\langle-1,1,0\rangle\).

- A translation adds a vector to a point's coordinates. - Multiply every component to find the second translation vector. - Apply the second translation to \(Q\), not to \(P\). - How could you check the final position using one combined translation?

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

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

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

1. \(2.5\mathbf{a}=\langle10,-5,10\rangle\) and \(6\mathbf{b}=\langle-6,18,6\rangle\). 2. \(\mathbf{c}=\langle10,-5,10\rangle+\langle-6,18,6\rangle=\langle4,13,16\rangle\). 3. \(\|\mathbf{c}\|=\sqrt{4^2+13^2+16^2}=\sqrt{441}=21\).

\(\mathbf{c}=\langle4,13,16\rangle\), \(\|\mathbf{c}\|=21\)

- 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 \(-4\): \(\mathbf{w}=\langle-4,12,-3\rangle\). 2. \(\|\mathbf{w}\|=\sqrt{(-4)^2+12^2+(-3)^2}=\sqrt{169}=13\).

\(\mathbf{w}=\langle-4,12,-3\rangle\), \(\|\mathbf{w}\|=13\)

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

1. In part a, \(\mathbf{a}=\langle5,-10,10\rangle\), so \(\|\mathbf{a}\|=\sqrt{5^2+(-10)^2+10^2}=15\). 2. In part b, factor the common vector: \(\mathbf{b}=(1.8-0.8)\langle0,4,-3\rangle=\langle0,4,-3\rangle\). 3. Thus, \(\|\mathbf{b}\|=\sqrt{4^2+(-3)^2}=5\).

a) \(\mathbf{a}=\langle5,-10,10\rangle\), \(\|\mathbf{a}\|=15\) b) \(\mathbf{b}=\langle0,4,-3\rangle\), \(\|\mathbf{b}\|=5\)

- Rewrite the decimal as a fraction. - Find a common denominator for all components. - After clearing denominators, make sure the integer components have no common factor greater than \(1\).

1. Rewrite \(-0.25\) as \(-\frac14\). 2. The least common denominator of \(6\) and \(4\) is \(12\). 3. Write the components with denominator \(12\): \(\frac56=\frac{10}{12}\), \(-\frac14=-\frac3{12}\), and \(1=\frac{12}{12}\). 4. Factor out \(\frac1{12}\): \(\mathbf{a}=\frac1{12}\langle10,-3,12\rangle\).

\(\mathbf{a}=\frac1{12}\langle10,-3,12\rangle\)

- 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 and second components give \(k=\frac16\), while the third gives \(k=\frac5{24}\). 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 \(12=(-1.5)(-8)\). 3. From the first components, \(a=2s=2(-1.5)=-3\).

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: \(\mathbf{v}=\langle2\cdot1-3\cdot1+3,2\cdot0-3(-1)+0,2\cdot4-3\cdot2+4\rangle=\langle2,3,6\rangle\). 2. Find the magnitude: \(\|\mathbf{v}\|=\sqrt{2^2+3^2+6^2}=\sqrt{49}=7\).

\(\mathbf{v}=\langle2,3,6\rangle\); \(\|\mathbf{v}\|=7\)

- 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: \(3\langle-1,2,1\rangle=\langle-3,6,3\rangle\), \(2\langle3,0,4\rangle=\langle6,0,8\rangle\), and \(4\langle0.5,0.5,0.75\rangle=\langle2,2,3\rangle\). 2. Combine the results: \(\mathbf{w}=\langle-3+6-2,6+0-2,3+8-3\rangle=\langle1,4,8\rangle\). 3. Find the magnitude: \(\|\mathbf{w}\|=\sqrt{1^2+4^2+8^2}=\sqrt{81}=9\).

\(\mathbf{w}=\langle1,4,8\rangle\); \(\|\mathbf{w}\|=9\)

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

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

- 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: \(\langle1,-5,12\rangle+\langle5,13,12\rangle=\langle6,8,24\rangle\). 2. Multiply by \(0.5\): \(\mathbf{a}=\langle3,4,12\rangle\). 3. Find the magnitude: \(\|\mathbf{a}\|=\sqrt{3^2+4^2+12^2}=\sqrt{169}=13\).

\(\mathbf{a}=\langle3,4,12\rangle\); \(\|\mathbf{a}\|=13\)

- 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: \(\mathbf{b}=(1.4+0.6)\langle2,5,-14\rangle=2\langle2,5,-14\rangle\). 2. Thus, \(\mathbf{b}=\langle4,10,-28\rangle\). 3. Its magnitude is \(\|\mathbf{b}\|=\sqrt{4^2+10^2+(-28)^2}=\sqrt{900}=30\).

\(\mathbf{b}=\langle4,10,-28\rangle\); \(\|\mathbf{b}\|=30\)

- 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, \(x=3(-4)=-12\) and \(z=2(-4)=-8\).

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 \(-1\): \(-\mathbf{a}=\langle-7,2.5,-k\rangle\). 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, \(-\mathbf{v}=\langle-v_1,-v_2,-v_3\rangle\).

a) \(-\mathbf{a}=\langle-7,2.5,-k\rangle\) b) \(k=-4.5\) c) Negate every component.

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

1. Since \(\mathbf{v}\) points in the same direction and has twice the magnitude, \(\mathbf{v}=2\mathbf{u}=\langle2,-4,0\rangle\). 2. Add \(\mathbf{v}\) to each vertex: \(A'=(3, 0, -1)+\langle2,-4,0\rangle=(5, -4, -1)\), \(B'=(1, 4, 2)+\langle2,-4,0\rangle=(3, 0, 2)\), and \(C'=(-2, 2, 5)+\langle2,-4,0\rangle=(0, -2, 5)\).

a) \(\mathbf{v}=\langle2,-4,0\rangle\) b) \(A'=(5, -4, -1)\), \(B'=(3, 0, 2)\), \(C'=(0, -2, 5)\)

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

1. \(\mathbf{v}=Q-P=\langle3,-3,6\rangle\). 2. \(3\mathbf{v}=\langle9,-9,18\rangle\). 3. \(R=P+3\mathbf{v}=(2, 0, -5)+\langle9,-9,18\rangle=(11, -9, 13)\).

1. \(\mathbf{v}=\langle3,-3,6\rangle\) 2. \(R=(11, -9, 13)\)

- Form the displacement vector from the starting point to the sensor. - One scalar must multiply every component of \(\mathbf{u}\) to obtain that displacement. - Interpret the sign of the scalar geometrically after finding it.

1. \(\overrightarrow{LS}=\langle-1-2,7-1,8-5\rangle=\langle-3,6,3\rangle\). 2. \(\langle-3,6,3\rangle=3\langle-1,2,1\rangle=3\mathbf{u}\). 3. The scalar is \(3\), which is nonnegative, so the displacement is allowed. A negative scalar would point in the direction opposite \(\mathbf{u}\).

Yes. \(\overrightarrow{LS}=3\mathbf{u}\), and \(3\ge0\), so the displacement is in the allowed direction.

- 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 can be written using the zero vector as its starting point and a nonzero direction vector.

1. Points on the same line through the origin have position vectors that are scalar multiples of one nonzero direction vector. 2. Using \(\langle3,-2,4\rangle\), we have \(\overrightarrow{OA}=3\langle3,-2,4\rangle\), \(\overrightarrow{OB}=-\langle3,-2,4\rangle\), and \(\overrightarrow{OD}=-\frac{1}{2}\langle3,-2,4\rangle\). 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 \(g:\mathbf{x}=t\langle3,-2,4\rangle\).

1. Points \(A\), \(B\), and \(D\) lie on the same line through the origin; \(C\) does not. 2. \(g:\mathbf{x}=t\langle3,-2,4\rangle\), where \(t\in\mathbb{R}\).

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

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

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

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

1. \(\overrightarrow{AB}=\langle4,2,2\rangle\) and \(\overrightarrow{DC}=C-D=\langle2,1,1\rangle\). 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. \(\|\mathbf{u}\|=\sqrt{3^2+(-4)^2+12^2}=\sqrt{169}=13\). 2. \(\mathbf{v}=-2\mathbf{u}=\langle-6,8,-24\rangle\), so \(\|\mathbf{v}\|=\sqrt{(-6)^2+8^2+(-24)^2}=\sqrt{676}=26\). 3. Alternatively, \(\|\mathbf{v}\|=|-2|\|\mathbf{u}\|=2\cdot13=26\). 4. The negative sign reverses the direction, and the absolute value \(2\) doubles the magnitude.

a) \(13\) b) From \(\mathbf{v}=\langle-6,8,-24\rangle\), \(\|\mathbf{v}\|=\sqrt{36+64+576}=26\). Also, \(\|\mathbf{v}\|=|-2|\|\mathbf{u}\|=2\cdot13=26\). c) The direction reverses and the magnitude doubles.

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

1. \(\|\mathbf{a}\|=\sqrt{4^2+0^2+(-3)^2}=5\). 2. Since \(\|\mathbf{b}\|=|r|\|\mathbf{a}\|\), \(20=5|r|\), so \(|r|=4\). 3. Therefore, \(r=4\) or \(r=-4\). 4. For \(r=4\), \(\mathbf{b}\) points in the same direction as \(\mathbf{a}\). 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) If \(r=4\), \(\mathbf{b}\) points in the same direction as \(\mathbf{a}\); if \(r=-4\), it points in the opposite direction. Both have magnitude \(20\), four times the length of \(\mathbf{a}\).

- Form both displacement vectors from the same initial point. - The first two components determine the only possible scalar multiple. - Use the final component to determine the unknown coordinate, then interpret the scalar-multiple relationship geometrically.

1. \(\overrightarrow{AB}=\langle2,1,2\rangle\) and \(\overrightarrow{AC_k}=\langle4,2,k-1\rangle\). 2. The first two components require \(\overrightarrow{AC_k}=2\overrightarrow{AB}\). 3. Therefore, \(k-1=2\cdot2=4\), so \(k=5\). 4. When \(k=5\), \(\overrightarrow{AC_5}=2\overrightarrow{AB}\), so \(A\), \(B\), and \(C_5\) are collinear.

\(k=5\). The scalar is \(2\), and the three points are collinear.

- A scalar acts on the entire vector, not just one entry. - Check each coordinate of the student's result separately. - What does the sign of the scalar say about direction?

1. The student multiplied the first component by \(-3\) but left the second component unchanged. 2. Multiply both components: \(-3\langle2,-5\rangle=\langle-6,15\rangle\). 3. Because the scalar is negative, the correct vector points in the opposite direction.

Both components must be multiplied. The correct vector is \(\langle-6,15\rangle\), pointing in the opposite direction.

- Decide what relationship two dependent nonzero vectors must have. - Introduce one scalar multiplier that connects all corresponding components. - Use the repeated second and third components to express that multiplier. - Check both roots in the original vectors.

1. Two nonzero vectors are linearly dependent when one is a scalar multiple of the other. Let \(\mathbf{v}=k\mathbf{u}\). 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 \(a=-4\)

- Which points correspond to \(\lambda=0\) and \(\lambda=1\)? - Which parameter value gives the midpoint of \(\overline{AB}\)? - What happens to the point when the parameter is negative? - Think of \(\overrightarrow{AB}\) as a step from \(A\): where does a half-step or a backward step land?

1. The direction vector is \(\overrightarrow{AB}=\langle3,6,-3\rangle\). 2. For \(\lambda=0.4\), \(\mathbf{x}=\langle1,-2,3\rangle+0.4\langle3,6,-3\rangle=\langle2.2,0.4,1.8\rangle\). Thus, \(P=(2.2, 0.4, 1.8)\). 3. A point lies strictly between \(A\) and \(B\) when \(0<\lambda<1\). Its distance from \(A\) is \(\lambda\|\overrightarrow{AB}\|\), and its distance from \(B\) is \((1-\lambda)\|\overrightarrow{AB}\|\). The first is greater when \(\lambda>\frac12\). Therefore, \(\frac12<\lambda<1\). 4. When \(\lambda<0\), the point lies outside segment \(\overline{AB}\), beyond \(A\) on the side opposite \(B\).

a) \(P=(2.2, 0.4, 1.8)\) b) \(\lambda\in\left(\frac12,1\right)\) c) Point \(P\) lies beyond \(A\) on the extension of the line opposite \(B\).

- 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 scaled displacement to the starting point \(Q\).

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

a) \(\overrightarrow{PQ}=\langle-3,4,0\rangle\), \(\|\overrightarrow{PQ}\|=5\) b) \(R=(-7, 9, 1)\)

- Decompose each requested vector into a path along prism 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 \(D\to A\to B\to F\): \(\overrightarrow{DF}=\overrightarrow{DA}+\overrightarrow{AB}+\overrightarrow{BF}=-\mathbf{v}+\mathbf{u}+\mathbf{w}\). 2. Therefore, \(\overrightarrow{DF}=\mathbf{u}-\mathbf{v}+\mathbf{w}\). 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 \(\overline{FG}\), \(\overrightarrow{FM}=\frac12\overrightarrow{FG}=\frac12\mathbf{v}\). 5. Thus, \(\overrightarrow{AM}=\mathbf{w}+\mathbf{u}+\frac12\mathbf{v}=\mathbf{u}+\frac12\mathbf{v}+\mathbf{w}\).

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

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

1. In part a, \(\mathbf{c}=\langle3,3\sqrt2,-3\sqrt3\rangle\). Then \(\|\mathbf{c}\|=\sqrt{3^2+(3\sqrt2)^2+(-3\sqrt3)^2}=\sqrt{54}=3\sqrt6\). 2. In part b, \(\mathbf{d}=\langle2,1,-3\rangle+\langle1,2,9\rangle=\langle3,3,6\rangle\). Then \(\|\mathbf{d}\|=\sqrt{3^2+3^2+6^2}=\sqrt{54}=3\sqrt6\).

a) \(\mathbf{c}=\langle3,3\sqrt2,-3\sqrt3\rangle\), \(\|\mathbf{c}\|=3\sqrt6\) b) \(\mathbf{d}=\langle3,3,6\rangle\), \(\|\mathbf{d}\|=3\sqrt6\)

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

- 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 \(k=\frac{1}{4}\). 5. Therefore, \(\overrightarrow{OT}=\frac{3}{4}\langle2,-1,4\rangle+\frac{1}{4}\langle6,7,0\rangle=\langle3,1,3\rangle\). 6. Thus, \(T=(3,1,3)\).

a) \(\overrightarrow{OT}=\overrightarrow{OA}+k(\overrightarrow{OB}-\overrightarrow{OA})=(1-k)\overrightarrow{OA}+k\overrightarrow{OB}\) b) A \(1:3\) division gives \(k=\frac{1}{4}\), so \(\overrightarrow{OT}=\frac{3}{4}\langle2,-1,4\rangle+\frac{1}{4}\langle6,7,0\rangle=\langle3,1,3\rangle\) and \(T=(3,1,3)\).

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

a) \(\overrightarrow{OM_a}=\frac{1}{2}(\overrightarrow{OB}+\overrightarrow{OC})\) b) Since \(\overrightarrow{AG}=\frac{2}{3}\overrightarrow{AM_a}\), \(\overrightarrow{OG}=\overrightarrow{OA}+\frac{2}{3}(\overrightarrow{OM_a}-\overrightarrow{OA})=\frac{1}{3}\overrightarrow{OA}+\frac{2}{3}\cdot\frac{1}{2}(\overrightarrow{OB}+\overrightarrow{OC})=\frac{1}{3}(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})\).

- 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 \(\overrightarrow{DC}=-\frac{1}{4}\overrightarrow{CA}\), \(\overrightarrow{DE}=\frac{1}{4}(\overrightarrow{CB}-\overrightarrow{CA})\). 3. Because \(\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{CB}\), \(\overrightarrow{CB}-\overrightarrow{CA}=\overrightarrow{AB}\). 4. Therefore, \(\overrightarrow{DE}=\frac{1}{4}\overrightarrow{AB}\). 5. The positive scalar multiple means the vectors \(\overrightarrow{DE}\) and \(\overrightarrow{AB}\) have the same direction. Their segments are parallel, and \(DE\) is one-fourth as long as \(AB\).

\(\overrightarrow{DE}=\frac{1}{4}\overrightarrow{AB}\). Thus, \(\overline{DE}\parallel\overline{AB}\) and \(DE=\frac{1}{4}AB\). The directed vectors point the same way.

- Compute the magnitude of the base vector before considering any scalar multiple. - Write the coordinates of \(s\mathbf{d}\) explicitly for the coordinate-sum condition. - The magnitude of a scalar multiple depends on the absolute value of the scalar.

1. \(\|\mathbf{d}\|=\sqrt{3^2+(-2)^2+6^2}=\sqrt{49}=7\). 2. \(P_s=(3s,-2s,6s)\), so the coordinate sum is \(7s\). Thus, \(7s=21\), giving \(s=3\) and \(P_3=(9,-6,18)\). 3. \(\|s\mathbf{d}\|=|s|\|\mathbf{d}\|=7|s|\). Setting \(7|s|=35\) gives \(s=5\) or \(s=-5\). 4. The corresponding points are \((15,-10,30)\) and \((-15,10,-30)\).

a) \(7\) b) \(s=3\), so \(P_s=(9,-6,18)\) c) \(s=5\) or \(s=-5\), giving \((15,-10,30)\) and \((-15,10,-30)\)

- 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, \(m=3(-2)=-6\) and \(n=3\cdot4=12\). 3. The statement is false. For example, \(\mathbf{a}=\langle1,2,3\rangle\) and \(\mathbf{b}=\langle3,2,1\rangle\) 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, \(\langle1,2,3\rangle\) and \(\langle3,2,1\rangle\) 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. Adding gives \(2\overrightarrow{MN}=(\overrightarrow{MA}+\overrightarrow{MB})+\overrightarrow{AD}+\overrightarrow{BC}+(\overrightarrow{DN}+\overrightarrow{CN})\). 3. Since \(M\) and \(N\) are midpoints, \(\overrightarrow{MA}+\overrightarrow{MB}=\mathbf{0}\) and \(\overrightarrow{DN}+\overrightarrow{CN}=\mathbf{0}\). 4. Thus, \(2\overrightarrow{MN}=\overrightarrow{AD}+\overrightarrow{BC}\), and \(\overrightarrow{MN}=\frac{1}{2}(\overrightarrow{AD}+\overrightarrow{BC})\).

Following the two paths through \(A,D\) and \(B,C\) gives \(2\overrightarrow{MN}=(\overrightarrow{MA}+\overrightarrow{MB})+\overrightarrow{AD}+\overrightarrow{BC}+(\overrightarrow{DN}+\overrightarrow{CN})\). The first and last parenthesized sums are \(\mathbf{0}\), since \(M\) and \(N\) are midpoints. Hence \(\overrightarrow{MN}=\frac{1}{2}(\overrightarrow{AD}+\overrightarrow{BC})\).

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

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

- Express each midpoint as an average of endpoint position vectors. - Compare the difference of two midpoint vectors with a vector across the original quadrilateral. - Compare the two diagonal magnitudes.

1. Since \(\mathbf{e}=\frac{1}{2}(\mathbf{a}+\mathbf{b})\) and \(\mathbf{f}=\frac{1}{2}(\mathbf{b}+\mathbf{c})\), \(\overrightarrow{EF}=\mathbf{f}-\mathbf{e}=\frac{1}{2}(\mathbf{c}-\mathbf{a})=\frac{1}{2}\overrightarrow{AC}\). Likewise, \(\mathbf{h}=\frac{1}{2}(\mathbf{d}+\mathbf{a})\) and \(\mathbf{g}=\frac{1}{2}(\mathbf{c}+\mathbf{d})\), giving \(\overrightarrow{HG}=\frac{1}{2}\overrightarrow{AC}\). 2. The other opposite sides satisfy \(\overrightarrow{FG}=\overrightarrow{EH}=\frac{1}{2}\overrightarrow{BD}\). 3. \(\|\overrightarrow{AC}\|=\sqrt{6^2+0^2+8^2}=10\) and \(\|\overrightarrow{BD}\|=\sqrt{0^2+6^2+8^2}=10\). 4. Each side of \(EFGH\) has length \(5\), so its perimeter is \(20\). 5. Because all four sides are equal, \(EFGH\) is a rhombus.

1. \(\overrightarrow{EF}=\frac{1}{2}(\mathbf{b}+\mathbf{c})-\frac{1}{2}(\mathbf{a}+\mathbf{b})=\frac{1}{2}\overrightarrow{AC}\). Similarly, \(\overrightarrow{HG}=\frac{1}{2}(\mathbf{c}+\mathbf{d})-\frac{1}{2}(\mathbf{d}+\mathbf{a})=\frac{1}{2}\overrightarrow{AC}\). 2. Each side measures \(5\) units, since the two diagonals of \(ABCD\) both have length \(10\) units and each side of \(EFGH\) has half the corresponding diagonal's length. 3. The perimeter is \(20\) units. \(EFGH\) is a rhombus because its four sides have the same length.

- 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. \(\overrightarrow{PQ}=Q-P=\langle4,2,-2\rangle\), so \(S=R-\overrightarrow{PQ}=(1, 8, 2)\). 3. A \(2:1\) ratio from \(P\) means \(\overrightarrow{PT}=\frac23\overrightarrow{PR}\). 4. \(\overrightarrow{PR}=R-P=\langle3,9,-3\rangle\). 5. Therefore, \(T=P+\frac23\overrightarrow{PR}=(2, 1, 3)+\langle2,6,-2\rangle=(4, 7, 1)\).

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. \(\overrightarrow{AB}=B-A=\langle4,2,-4\rangle\), so \(\overrightarrow{DC}=\langle2,1,-2\rangle\). 3. Thus, \(D=C-\overrightarrow{DC}=(1, 8, 3)\). 4. The midpoint of \(\overline{AB}\) is \(M=(3, 3, 2)\). 5. \(\overrightarrow{MZ}=\langle0,3,-0.5\rangle\) and \(\overrightarrow{MC}=\langle0,6,-1\rangle\). 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. \(\overrightarrow{CG}=\langle-4,2,-2\rangle\), while \(\overrightarrow{GM}=\langle-2,1,-1\rangle\). 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. \(\overrightarrow{AB}=\langle4,4,-2\rangle\) and \(\overrightarrow{AC}=\langle-2,8,2\rangle\). 2. The midpoints are \(M=(5,3,3)\) and \(N=(2,5,5)\). 3. \(\overrightarrow{MN}=\langle-3,2,2\rangle\), while \(\overrightarrow{BC}=\langle-6,4,4\rangle\). Thus, \(\overrightarrow{MN}=\frac{1}{2}\overrightarrow{BC}\). 4. Generally, if \(\mathbf{m}=\frac{1}{2}(\mathbf{a}+\mathbf{b})\) and \(\mathbf{n}=\frac{1}{2}(\mathbf{a}+\mathbf{c})\), then \(\overrightarrow{MN}=\mathbf{n}-\mathbf{m}=\frac{1}{2}(\mathbf{c}-\mathbf{b})=\frac{1}{2}\overrightarrow{BC}\).

a) \(\overrightarrow{AB}=\langle4,4,-2\rangle\), \(\overrightarrow{AC}=\langle-2,8,2\rangle\) b) \(M=(5,3,3)\), \(N=(2,5,5)\) c) \(\overrightarrow{MN}=\langle-3,2,2\rangle=\frac{1}{2}\overrightarrow{BC}\) d) Let \(\mathbf{a},\mathbf{b},\mathbf{c}\) be the position vectors of the vertices. Then \(\mathbf{m}=\frac{1}{2}(\mathbf{a}+\mathbf{b})\) and \(\mathbf{n}=\frac{1}{2}(\mathbf{a}+\mathbf{c})\). Subtracting gives \(\overrightarrow{MN}=\mathbf{n}-\mathbf{m}=\frac{1}{2}(\mathbf{c}-\mathbf{b})=\frac{1}{2}\overrightarrow{BC}\).

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

1. Subtracting \(\mathbf{a}+\mathbf{d}\) from \(\mathbf{a}+\mathbf{c}=\mathbf{b}+\mathbf{d}\) gives \(\mathbf{c}-\mathbf{d}=\mathbf{b}-\mathbf{a}\). Hence \(\overrightarrow{AB}=\overrightarrow{DC}\). 2. \(\mathbf{m}_{AC}=\frac{1}{2}(\mathbf{a}+\mathbf{c})=\frac{1}{2}(\mathbf{b}+\mathbf{d})=\mathbf{m}_{BD}\). The diagonals therefore have the same midpoint. 3. A nondegenerate 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, \(\mathbf{v}_1=\langle v_1,v_2,-v_3\rangle\). 2. After reflection across the \(xz\)-plane, \(\mathbf{v}_2=\langle v_1,-v_2,-v_3\rangle\). 3. After reflection across the \(yz\)-plane, \(\mathbf{v}_{\text{final}}=\langle-v_1,-v_2,-v_3\rangle\). 4. Thus, \(\mathbf{v}_{\text{final}}=-\mathbf{v}\), 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) \(\mathbf{v}_1=\langle v_1,v_2,-v_3\rangle\), \(\mathbf{v}_2=\langle v_1,-v_2,-v_3\rangle\), and \(\mathbf{v}_{\text{final}}=\langle-v_1,-v_2,-v_3\rangle\) b) \(\mathbf{v}_{\text{final}}=-\mathbf{v}\), so \(k=-1\). c) The ray is returned toward the light source.

- 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.

- 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 \(\overrightarrow{OG}=\frac{1}{3}(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})\). 2. Since \(S\) is three-fourths of the way from \(D\) to \(G\), \(\overrightarrow{OS}=\overrightarrow{OD}+\frac{3}{4}(\overrightarrow{OG}-\overrightarrow{OD})\). 3. Substitution gives \(\overrightarrow{OS}=\frac{1}{4}(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD})\). 4. The edge midpoints satisfy \(\overrightarrow{OM_{AB}}=\frac{1}{2}(\overrightarrow{OA}+\overrightarrow{OB})\) and \(\overrightarrow{OM_{CD}}=\frac{1}{2}(\overrightarrow{OC}+\overrightarrow{OD})\). 5. Their midpoint has position vector \(\frac{1}{2}(\overrightarrow{OM_{AB}}+\overrightarrow{OM_{CD}})=\frac{1}{4}(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD})=\overrightarrow{OS}\). Therefore, it is \(S\).

1. \(\overrightarrow{OS}=\frac{1}{4}(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD})\) 2. \(\overrightarrow{OM_{AB}}=\frac{1}{2}(\overrightarrow{OA}+\overrightarrow{OB})\) and \(\overrightarrow{OM_{CD}}=\frac{1}{2}(\overrightarrow{OC}+\overrightarrow{OD})\). Thus, \(\frac{1}{2}(\overrightarrow{OM_{AB}}+\overrightarrow{OM_{CD}})=\overrightarrow{OS}\), proving that \(S\) is their midpoint.

- Rewrite each vector from \(P\) as a difference of position vectors. - Group the coefficients of \(B\) and \(C\) so they define a point on \(\overline{BC}\). - Can the resulting point be written as a weighted average of two points in more than one way?

1. Replace each displacement vector with a difference of position vectors: \(3(\overrightarrow{OA}-\overrightarrow{OP})+2(\overrightarrow{OB}-\overrightarrow{OP})+(\overrightarrow{OC}-\overrightarrow{OP})=\mathbf{0}\). 2. Combining terms gives \(6\overrightarrow{OP}=3\overrightarrow{OA}+2\overrightarrow{OB}+\overrightarrow{OC}\). 3. Hence \(\overrightarrow{OP}=\frac{1}{2}\overrightarrow{OA}+\frac{1}{3}\overrightarrow{OB}+\frac{1}{6}\overrightarrow{OC}\). 4. Choose \(M\) with \(\overrightarrow{OM}=\frac{2}{3}\overrightarrow{OB}+\frac{1}{3}\overrightarrow{OC}\). This places \(M\) on \(\overline{BC}\) with \(BM:MC=1:2\). 5. Now \(\overrightarrow{OP}=\frac{1}{2}\overrightarrow{OA}+\frac{1}{2}\overrightarrow{OM}\), so \(P\) is the midpoint of \(\overline{AM}\) and \(AP:PM=1:1\).

1. From \(6\overrightarrow{OP}=3\overrightarrow{OA}+2\overrightarrow{OB}+\overrightarrow{OC}\), \(\overrightarrow{OP}=\frac{1}{2}\overrightarrow{OA}+\frac{1}{3}\overrightarrow{OB}+\frac{1}{6}\overrightarrow{OC}\). 2. Set \(\overrightarrow{OM}=\frac{2}{3}\overrightarrow{OB}+\frac{1}{3}\overrightarrow{OC}\). Then \(BM:MC=1:2\) and \(\overrightarrow{OP}=\frac{1}{2}(\overrightarrow{OA}+\overrightarrow{OM})\). Therefore, \(P\) is the midpoint of \(\overline{AM}\) 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 position vectors are \(\mathbf{p}=\frac{1}{2}(\mathbf{a}+\mathbf{b})\) and \(\mathbf{q}=\frac{1}{2}(\mathbf{b}+\mathbf{c})\). Thus, \(\overrightarrow{PQ}=\mathbf{q}-\mathbf{p}=\frac{1}{2}(\mathbf{c}-\mathbf{a})\). 2. Likewise, \(\mathbf{s}=\frac{1}{2}(\mathbf{a}+\mathbf{d})\) and \(\mathbf{r}=\frac{1}{2}(\mathbf{c}+\mathbf{d})\), so \(\overrightarrow{SR}=\mathbf{r}-\mathbf{s}=\frac{1}{2}(\mathbf{c}-\mathbf{a})=\overrightarrow{PQ}\). A nondegenerate quadrilateral with equal opposite-side vectors is a parallelogram. 3. The equality \(\overrightarrow{SR}=\overrightarrow{PQ}\) means \(R=S+Q-P\). Hence the four points lie in the plane through \(P,Q,S\), if these are noncollinear, or on a common line when they are collinear. 4. When \(A=C\), \(\overrightarrow{PQ}=\mathbf{0}\) and \(\overrightarrow{SR}=\mathbf{0}\), so \(P=Q\) and \(S=R\). The figure reduces to a segment, or a point if the two remaining positions coincide.

1. \(\overrightarrow{PQ}=\frac{1}{2}(\mathbf{b}+\mathbf{c})-\frac{1}{2}(\mathbf{a}+\mathbf{b})=\frac{1}{2}(\mathbf{c}-\mathbf{a})\). 2. \(\overrightarrow{SR}=\frac{1}{2}(\mathbf{c}+\mathbf{d})-\frac{1}{2}(\mathbf{a}+\mathbf{d})=\frac{1}{2}(\mathbf{c}-\mathbf{a})=\overrightarrow{PQ}\). Equal opposite-side vectors give a parallelogram unless the points coincide or are collinear. 3. Since \(R=S+Q-P\), the four points lie in a common plane, including when they are collinear. 4. If \(A=C\), then \(P=Q\) and \(S=R\); the figure becomes a segment or a point.