Free math worksheets

- Vector addition combines corresponding components. - Add the horizontal components together. - Add the vertical components together.

1. Add corresponding components: \(\mathbf{u}+\mathbf{v}=\langle2+(-5),-3+4\rangle\). 2. Therefore, \(\mathbf{u}+\mathbf{v}=\langle-3,1\rangle\).

\(\langle-3,1\rangle\)

- Follow the direction from \(O\) to \(A\), then from \(A\) to \(B\). - The sum runs directly from the starting point of the first vector to the endpoint of the second. - Read the final endpoint relative to the origin.

1. By the tip-to-tail rule, \(\mathbf{u}+\mathbf{v}\) is the displacement from \(O\) to \(B\). 2. The diagram places \(B\) at \((4, 2)\), so \(\mathbf{u}+\mathbf{v}=\langle4,2\rangle\).

\(\langle4,2\rangle\)

- Read the terminal-point coordinates of \(\mathbf{u}\) and \(\mathbf{v}\). - Subtract corresponding components. - Compare the result with the displacement from \(B\) to \(A\).

1. From the diagram, \(\mathbf{u}=\langle4,1\rangle\) and \(\mathbf{v}=\langle1,3\rangle\). 2. Subtract components: \(\mathbf{u}-\mathbf{v}=\langle4-1,1-3\rangle=\langle3,-2\rangle.\) 3. Since \(\overrightarrow{BA}=A-B=\langle3,-2\rangle\), the difference is represented by the vector directed from \(B\) to \(A\).

\(\mathbf{u}-\mathbf{v}=\langle3,-2\rangle\), represented by \(\overrightarrow{BA}\).

- Vector subtraction is componentwise. - Be careful when subtracting a negative first component.

1. Subtract corresponding components: \(\mathbf{u}-\mathbf{v}=\langle2-(-1),-5-3,1-4\rangle.\) 2. Therefore, \(\mathbf{u}-\mathbf{v}=\langle3,-8,-3\rangle\).

\(\langle3,-8,-3\rangle\).

- Add or subtract corresponding real and imaginary parts. - A complex number \(a+bi\) corresponds to the point \((a, b)\). - Use the magnitude formula for the difference. - Think about what the length of the segment between two points represents.

1. \(s=(4-2)+(2+3)i=2+5i\). 2. \(d=(4-(-2))+(2-3)i=6-i\). 3. The endpoint for \(s\) is \((2, 5)\), and the endpoint for \(d\) is \((6, -1)\). 4. \(|z_1-z_2|=|6-i|=\sqrt{6^2+(-1)^2}=\sqrt{37}\approx6.08\). 5. The magnitude of the difference is the distance between the points represented by \(z_1\) and \(z_2\).

a) \(s=2+5i\); \(d=6-i\) b) \((2, 5)\) and \((6, -1)\) c) \(|z_1-z_2|=\sqrt{37}\approx6.08\); this is the distance between the two points.

- Replace a subtracted vector by the vector with reversed endpoints. - Use head-to-tail addition when one vector ends where the next begins. - You may reorder a vector sum. - Distribute a negative sign across parentheses carefully.

1. In part a), \(\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\), and \(\overrightarrow{AC}+\overrightarrow{CD}=\overrightarrow{AD}\). 2. In part b), \(-\overrightarrow{RQ}=\overrightarrow{QR}\). Therefore, \(\overrightarrow{PQ}+\overrightarrow{QR}+\overrightarrow{RS}=\overrightarrow{PR}+\overrightarrow{RS}=\overrightarrow{PS}\). 3. In part c), \(\overrightarrow{XY}-\overrightarrow{ZY}-\overrightarrow{XZ} =\overrightarrow{XY}+\overrightarrow{YZ}+\overrightarrow{ZX}\). Then \(\overrightarrow{XY}+\overrightarrow{YZ}=\overrightarrow{XZ}\), so the result is \(\overrightarrow{XZ}+\overrightarrow{ZX}=\mathbf{0}\).

a) \(\overrightarrow{AD}\) b) \(\overrightarrow{PS}\) c) \(\mathbf{0}\)

- Evaluate the brackets first. - Add and subtract vectors componentwise. - Use the three-dimensional magnitude formula. - Watch the signs when subtracting negative components.

1. Add the vectors inside the brackets: \(\langle 4,1,5\rangle+\langle -2,-7,10\rangle=\langle 2,-6,15\rangle\). 2. Subtract componentwise: \(\mathbf{v}=\langle 8,-2,3\rangle-\langle 2,-6,15\rangle=\langle 6,4,-12\rangle\). 3. \(\|\mathbf{v}\|=\sqrt{6^2+4^2+(-12)^2}=\sqrt{196}=14\).

\(\mathbf{v}=\langle 6,4,-12\rangle\), \(\|\mathbf{v}\|=14\)

- Find each displacement vector by subtracting endpoint coordinates. - Add vectors componentwise. - Use the three-dimensional magnitude formula.

1. \(\overrightarrow{AB}=B-A=\langle 5,2,-1\rangle\) and \(\overrightarrow{AC}=C-A=\langle 3,2,2\rangle\). 2. Add componentwise: \(\mathbf{s}=\langle 5,2,-1\rangle+\langle 3,2,2\rangle=\langle 8,4,1\rangle\). 3. \(\|\mathbf{s}\|=\sqrt{8^2+4^2+1^2}=\sqrt{81}=9\).

\(\mathbf{s}=\langle 8,4,1\rangle\), \(\|\mathbf{s}\|=9\)

- Use the stated order \(A,B,C,D\). - Opposite sides of a parallelogram are represented by equal vectors. - Find \(\overrightarrow{BC}\), then translate \(A\) by that vector. - Check that \(\overrightarrow{AB}=\overrightarrow{DC}\).

1. In parallelogram \(ABCD\), \(\overrightarrow{AD}=\overrightarrow{BC}\). 2. \(\overrightarrow{BC}=C-B=\langle -6,6,-2\rangle\). 3. Add this vector to \(A\): \(D=A+\overrightarrow{BC}=(1,5,-2)+(-6,6,-2)=(-5,11,-4)\).

\(D=(-5,11,-4)\)

- Use the stated vertex order. - Opposite sides of a parallelogram are represented by equal vectors. - Solve \(\overrightarrow{SR}=\overrightarrow{PQ}\) for \(S\).

1. \(\overrightarrow{PQ}=Q-P=\langle 4,3,0\rangle\). 2. In parallelogram \(PQRS\), \(\overrightarrow{SR}=\overrightarrow{PQ}\). 3. Therefore, \(S=R-\overrightarrow{PQ}=(5,6,0)-(4,3,0)=(1,3,0)\).

\(S=(1,3,0)\)

- Break each vector into a path along known edges. - Reverse the sign when traversing an edge opposite its defined direction. - Opposite edges of a parallelepiped represent equal vectors.

1. Follow the path \(H\to D\to A\to B\): \(\overrightarrow{HB}=\overrightarrow{HD}+\overrightarrow{DA}+\overrightarrow{AB}\). 2. Since \(\overrightarrow{HD}=-\mathbf{c}\), \(\overrightarrow{DA}=-\mathbf{b}\), and \(\overrightarrow{AB}=\mathbf{a}\), \(\overrightarrow{HB}=\mathbf{a}-\mathbf{b}-\mathbf{c}\). 3. Follow the path \(E\to F\to G\): \(\overrightarrow{EG}=\overrightarrow{EF}+\overrightarrow{FG}\). 4. Opposite edges of a parallelepiped are equal and parallel, so \(\overrightarrow{EF}=\mathbf{a}\) and \(\overrightarrow{FG}=\mathbf{b}\). Therefore, \(\overrightarrow{EG}=\mathbf{a}+\mathbf{b}\).

\(\overrightarrow{HB}=\mathbf{a}-\mathbf{b}-\mathbf{c}\) \(\overrightarrow{EG}=\mathbf{a}+\mathbf{b}\)

- Distinguish carefully between vectors and scalars. - Check what type of object each operation should produce. - Recall the commutative and distributive properties. - Vector magnitude uses the absolute value of a scalar multiplier.

1. Statement a) is true by the commutative property of vector addition. 2. Statement b) is false because the result is a vector: \(\mathbf{a}+\mathbf{a}=2\mathbf{a}\). 3. Statement c) is true because \(\overrightarrow{QP}=-\overrightarrow{PQ}\), so their sum is the zero vector. 4. Statement d) is true by the distributive property. 5. Statement e) is true because \(\lVert k\mathbf{v}\rVert=|k|\lVert\mathbf{v}\rVert\).

a) True. b) False; \(\mathbf{a}+\mathbf{a}=2\mathbf{a}\). c) True. d) True. e) True.

- Compute terminal point minus initial point. - Reversing endpoints negates a vector but preserves its magnitude. - Use the magnitude formula. - Add corresponding components to verify the vector equation.

1. \(\overrightarrow{AB}=B-A=\langle 3,4,0\rangle\), and \(\overrightarrow{BA}=-\overrightarrow{AB}=\langle -3,-4,0\rangle\). 2. Both vectors have magnitude \(\sqrt{3^2+4^2}=5\). 3. \(\overrightarrow{AC}=C-A=\langle 3,0,-4\rangle\), with magnitude \(5\). 4. \(\overrightarrow{BC}=C-B=\langle 0,-4,-4\rangle\). Therefore, \(\overrightarrow{AB}+\overrightarrow{BC}=\langle 3,4,0\rangle+\langle 0,-4,-4\rangle=\langle 3,0,-4\rangle=\overrightarrow{AC}\).

a) \(\overrightarrow{AB}=\langle 3,4,0\rangle\), \(\overrightarrow{BA}=\langle -3,-4,0\rangle\), and both magnitudes are \(5\). b) \(\overrightarrow{AC}=\langle 3,0,-4\rangle\), with magnitude \(5\). c) \(\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\).

- Compute each displacement as terminal point minus initial point. - Add corresponding components. - Compare the beginning and ending letters of the vector chain.

1. \(\overrightarrow{AB}=B-A=\langle 4,4,-3\rangle\), and \(\overrightarrow{BC}=C-B=\langle -7,3,6\rangle\). 2. \(\mathbf{s}=\langle 4,4,-3\rangle+\langle -7,3,6\rangle=\langle -3,7,3\rangle\). 3. \(\overrightarrow{AC}=C-A=\langle -3,7,3\rangle\). Therefore, \(\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\). Geometrically, traveling from \(A\) to \(B\) and then from \(B\) to \(C\) has the same net displacement as traveling directly from \(A\) to \(C\).

1. \(\overrightarrow{AB}=\langle 4,4,-3\rangle\), \(\overrightarrow{BC}=\langle -7,3,6\rangle\) 2. \(\mathbf{s}=\langle -3,7,3\rangle\) 3. \(\overrightarrow{AC}=\langle -3,7,3\rangle\), so \(\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\).

- Reverse the endpoints to rewrite a negative vector. - A vector chain can be simplified repeatedly. - A closed path has zero net displacement.

1. \(\overrightarrow{PQ}+\overrightarrow{QR}=\overrightarrow{PR}\), and \(\overrightarrow{PR}+\overrightarrow{RS}=\overrightarrow{PS}\). 2. Since \(-\overrightarrow{CB}=\overrightarrow{BC}\), \(\overrightarrow{AB}-\overrightarrow{CB} =\overrightarrow{AB}+\overrightarrow{BC} =\overrightarrow{AC}\). Thus, \(\mathbf{x}=\overrightarrow{AC}\). 3. The vectors trace a closed path from \(A\) back to \(A\), so their sum is \(\mathbf{0}\).

1. \(\overrightarrow{PS}\) 2. \(\mathbf{x}=\overrightarrow{AC}\) 3. \(\mathbf{0}\); the path begins and ends at \(A\).

- Use terminal point minus initial point. - The opposite vector has all component signs reversed. - Reversing the endpoints reverses the vector.

1. \(\overrightarrow{PQ}=Q-P=\langle 4,6,-4\rangle\). 2. \(-\mathbf{v}=\langle -4,-6,4\rangle\). 3. \(\overrightarrow{QP}=P-Q=\langle -4,-6,4\rangle\). 4. Therefore, \(-\overrightarrow{PQ}=\overrightarrow{QP}\).

a) \(\overrightarrow{PQ}=\langle 4,6,-4\rangle\) b) \(-\mathbf{v}=\langle -4,-6,4\rangle\) c) \(\overrightarrow{QP}=\langle -4,-6,4\rangle\), so \(-\overrightarrow{PQ}=\overrightarrow{QP}\).

- Follow the closed path \(A\to B\to C\to A\). - Reverse endpoints to find an opposite vector. - Rewrite subtraction as addition of the opposite. - Reorder the sum to form a head-to-tail chain.

1. \(\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\). 2. Therefore, \(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA} =\overrightarrow{AC}+\overrightarrow{CA} =\mathbf{0}\). 3. For part b), \(\overrightarrow{AC}-\overrightarrow{AB} =\overrightarrow{AC}+\overrightarrow{BA}\). 4. Reorder the sum and apply the triangle rule: \(\overrightarrow{BA}+\overrightarrow{AC} =\overrightarrow{BC}\). Thus, \(\mathbf{x}=\overrightarrow{BC}\).

a) \(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\mathbf{0}\) b) \(\mathbf{x}=\overrightarrow{BC}\)

- Use paths that pass through \(A\). - Reversing a vector changes its sign. - Draw two vectors with a common initial point and connect their terminal points.

1. \(\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}=-\mathbf{u}+\mathbf{v}=\mathbf{v}-\mathbf{u}\). 2. \(\overrightarrow{CD}=\overrightarrow{CA}+\overrightarrow{AD}=-\mathbf{v}+\mathbf{w}=\mathbf{w}-\mathbf{v}\). 3. \(\overrightarrow{DB}=\overrightarrow{DA}+\overrightarrow{AB}=-\mathbf{w}+\mathbf{u}=\mathbf{u}-\mathbf{w}\). 4. When \(\mathbf{a}\) and \(\mathbf{b}\) share an initial point, \(\mathbf{b}-\mathbf{a}\) is the vector from the terminal point of \(\mathbf{a}\) to the terminal point of \(\mathbf{b}\).

a) \(\overrightarrow{BC}=\mathbf{v}-\mathbf{u}\), \(\overrightarrow{CD}=\mathbf{w}-\mathbf{v}\), and \(\overrightarrow{DB}=\mathbf{u}-\mathbf{w}\) b) \(\mathbf{b}-\mathbf{a}\) points from the terminal point of \(\mathbf{a}\) to the terminal point of \(\mathbf{b}\).

- A translation adds a vector to a point. - Apply the second translation to \(B\). - Successive translations combine by vector addition.

1. \(B=A+\mathbf{v}=(7,-2,4)+(-3,5,2)=(4,3,6)\). 2. \(C=B+\mathbf{w}=(4,3,6)+(1,-4,6)=(5,-1,12)\). 3. The combined translation is \(\mathbf{u}=\mathbf{v}+\mathbf{w}=\langle -2,1,8\rangle\).

1. \(B=(4,3,6)\) 2. \(C=(5,-1,12)\) 3. \(\mathbf{u}=\langle -2,1,8\rangle\)

- Write start plus displacement equals destination. - Reverse the displacement by subtraction. - Watch the sign when subtracting a negative component. - Check by adding the displacement back to your result.

1. The translation equation is \(P+\mathbf{v}=L\). 2. Solve for the starting point: \(P=L-\mathbf{v}\). 3. \(P=(4,15,20)-(12,-8,5)=(-8,23,15)\).

\(P=(-8,23,15)\)

- Combine the two translations by adding their vectors. - Reverse a translation by subtracting its vector. - Write original plus total displacement equals final. - Check by applying both translations to your result.

1. The combined translation is \(\mathbf{s}=\mathbf{u}+\mathbf{w}=\langle 2,3,-4\rangle\). 2. Reverse the combined translation: \(A=E-\mathbf{s}\). 3. \(A=(1,7,-2)-(2,3,-4)=(-1,4,2)\).

\(A=(-1,4,2)\)

- Find a translation vector as image minus original. - Apply the same vector to \(C\). - Reversing a translation negates its vector. - Check that the forward and reverse vectors sum to \(\mathbf{0}\).

1. \(\mathbf{v}=B-A=\langle -3,8,-5\rangle\). 2. \(D=C+\mathbf{v}=(-3,2,1)+(-3,8,-5)=(-6,10,-4)\). 3. The reverse translation uses the opposite vector: \(\mathbf{u}=A-B=\langle 3,-8,5\rangle=-\mathbf{v}\).

a) \(\mathbf{v}=\langle -3,8,-5\rangle\) b) \(D=(-6,10,-4)\) c) \(\mathbf{u}=\langle 3,-8,5\rangle=-\mathbf{v}\)

- Compare a point on the first base with its corresponding point on the second base. - Apply the same translation vector to every vertex. - The length of a lateral edge equals the magnitude of the translation vector.

1. Subtract the coordinates of \(A\) from those of \(A'\): \(\mathbf{v}=\overrightarrow{AA'}=\langle 0,0,8\rangle\). 2. Translate \(B\) and \(C\) by \(\mathbf{v}\): \(B'=(6,1,9)\) and \(C'=(2,4,9)\). 3. A lateral edge has length \(\|\mathbf{v}\|=\sqrt{0^2+0^2+8^2}=8\).

a) \(\mathbf{v}=\langle 0,0,8\rangle\) b) \(B'(6,1,9)\) and \(C'(2,4,9)\) c) \(8\) units

- Compute terminal point minus initial point. - Reversing a directed segment negates its vector. - Compare the signs of corresponding components.

1. \(\overrightarrow{RS}=S-R=\langle 3.5,-7.5,6\rangle\). 2. \(\overrightarrow{SR}=R-S=\langle -3.5,7.5,-6\rangle\). 3. Every component is negated, so \(\overrightarrow{SR}=-\overrightarrow{RS}\). The vectors have the same magnitude and opposite directions.

a) \(\overrightarrow{RS}=\langle 3.5,-7.5,6\rangle\) b) \(\overrightarrow{SR}=\langle -3.5,7.5,-6\rangle\) c) \(\overrightarrow{SR}=-\overrightarrow{RS}\); they have equal magnitude and opposite directions.

- Add and subtract vectors component by component. - Multiply every component of a vector by the scalar. - Pay close attention to signs when subtracting a negative component.

1. For part a), combine corresponding components: \(\mathbf{u}=\langle 4+(-3)-1,-2+5-1,7+0-(-2)\rangle=\langle 0,2,9\rangle\). 2. For part b), first multiply \(\mathbf{c}\) by \(3\): \(3\mathbf{c}=\langle 3,3,-6\rangle\). 3. Then add \(\mathbf{a}\): \(\mathbf{v}=\langle 7,1,1\rangle\).

a) \(\mathbf{u}=\langle 0,2,9\rangle\) b) \(\mathbf{v}=\langle 7,1,1\rangle\)

- Distribute the subtraction sign as you would in an algebraic expression. - What does \(\mathbf{b}-\mathbf{a}\) represent when \(\mathbf{a}\) and \(\mathbf{b}\) are position vectors? - Opposite vectors have equal magnitudes and opposite directions.

1. Distribute the subtraction: \(\mathbf{b}-(\mathbf{b}-\mathbf{a})=\mathbf{b}-\mathbf{b}+\mathbf{a}=\mathbf{a}\). 2. Since \(\mathbf{b}-\mathbf{a}=\overrightarrow{AB}\), subtracting \(\overrightarrow{AB}\) from the position vector of \(B\) returns the position vector of \(A\). 3. The vector \(\mathbf{a}-\mathbf{b}=\overrightarrow{BA}\). Its opposite vector points from \(A\) to \(B\), so \(-(\mathbf{a}-\mathbf{b})=\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\).

1. \(\mathbf{a}\) 2. Starting at \(B\) and subtracting \(\overrightarrow{AB}\) returns to \(A\). 3. \(\mathbf{a}-\mathbf{b}\) points from \(B\) to \(A\), so its opposite is \(\mathbf{b}-\mathbf{a}\), which points from \(A\) to \(B\).

- Evaluate the expression inside the parentheses first. - Subtract and add corresponding components. - Be careful when subtracting a negative number.

1. Subtract inside the parentheses: \(\langle 8,-4,7\rangle-\langle -5,2,11\rangle=\langle 13,-6,-4\rangle\). 2. Add the remaining vector: \(\langle -12,5,-3\rangle+\langle 13,-6,-4\rangle=\langle 1,-1,-7\rangle\).

\(\langle 1,-1,-7\rangle\)

- Evaluate the two parenthesized vector expressions separately. - Work one component at a time. - Watch the signs when subtracting the second result.

1. Evaluate the first pair of parentheses: \(\langle 4.2,-1.8,0\rangle-\langle -0.8,2.5,3.7\rangle=\langle 5.0,-4.3,-3.7\rangle\). 2. Evaluate the second pair: \(\langle 5.5,-6,1.2\rangle+\langle -2.1,1.7,-4.9\rangle=\langle 3.4,-4.3,-3.7\rangle\). 3. Subtract the results: \(\langle 5.0,-4.3,-3.7\rangle-\langle 3.4,-4.3,-3.7\rangle=\langle 1.6,0,0\rangle\).

\(\langle 1.6,0,0\rangle\)

- A displacement vector is found by subtracting initial coordinates from terminal coordinates. - Translation by a vector means adding that vector to the position vector of the point. - For the final check, compare two displacement vectors rather than using a line equation.

1. \(\overrightarrow{AB}=B-A=\langle 3,2,2\rangle\). 2. \(\overrightarrow{OP}=\langle 6,4,4\rangle=2\langle 3,2,2\rangle=2\overrightarrow{AB}\). 3. \(\overrightarrow{OP'}=\overrightarrow{OP}+\overrightarrow{OA}=\langle 6,4,4\rangle+\langle 2,-1,5\rangle=\langle 8,3,9\rangle\), so \(P'=(8,3,9)\). 4. \(\overrightarrow{AP'}=P'-A=\langle 6,4,4\rangle=\overrightarrow{OP}\).

a) \(\langle 3,2,2\rangle\) b) \(\overrightarrow{OP}=2\overrightarrow{AB}\) c) \(P'=(8,3,9)\) d) \(\overrightarrow{AP'}=\langle 6,4,4\rangle=\overrightarrow{OP}\)

- Treat \(\mathbf{x}\) as an unknown in an algebraic equation. - Distribute each scalar carefully, including the negative sign. - Collect all terms containing \(\mathbf{x}\) on one side.

1. Distribute on both sides: \(2\mathbf{x}-\langle 8,-2,6\rangle=\langle 10,0,-5\rangle-\langle 0,6,3\rangle+3\mathbf{x}\). 2. Combine the constant vectors: \(2\mathbf{x}-\langle 8,-2,6\rangle=3\mathbf{x}+\langle 10,-6,-8\rangle\). 3. Isolate \(\mathbf{x}\): \(-\langle 8,-2,6\rangle=\mathbf{x}+\langle 10,-6,-8\rangle\). 4. Therefore, \(\mathbf{x}=\langle -8,2,-6\rangle-\langle 10,-6,-8\rangle=\langle -18,8,2\rangle\).

\(\mathbf{x}=\langle -18,8,2\rangle\)

- Build a path from each initial point to its terminal point. - Use the head-to-tail rule for vector addition. - Reverse a vector's sign only when traversing it backward.

1. \(\overrightarrow{KM}=\overrightarrow{KL}+\overrightarrow{LM}=\mathbf{u}+\mathbf{v}\). 2. \(\overrightarrow{NM}=\overrightarrow{NK}+\overrightarrow{KL}+\overrightarrow{LM}=\mathbf{w}+\mathbf{u}+\mathbf{v}\). 3. \(\overrightarrow{KO}=\overrightarrow{KL}+\overrightarrow{LM}+\overrightarrow{MO}=\mathbf{u}+\mathbf{v}+\mathbf{x}\). 4. \(\overrightarrow{LO}=\overrightarrow{LM}+\overrightarrow{MO}=\mathbf{v}+\mathbf{x}\). 5. \(\overrightarrow{NO}=\overrightarrow{NK}+\overrightarrow{KL}+\overrightarrow{LM}+\overrightarrow{MO}=\mathbf{w}+\mathbf{u}+\mathbf{v}+\mathbf{x}\).

a) \(\overrightarrow{KM}=\mathbf{u}+\mathbf{v}\) b) \(\overrightarrow{NM}=\mathbf{w}+\mathbf{u}+\mathbf{v}\) c) \(\overrightarrow{KO}=\mathbf{u}+\mathbf{v}+\mathbf{x}\) d) \(\overrightarrow{LO}=\mathbf{v}+\mathbf{x}\) e) \(\overrightarrow{NO}=\mathbf{w}+\mathbf{u}+\mathbf{v}+\mathbf{x}\)

- Write each diagonal midpoint as the average of its endpoint vectors. - Equal midpoints characterize diagonals that bisect each other. - Recall the parallelogram rule for vector addition.

1. The midpoint of \(\overline{OB}\) has position vector \(\frac12\mathbf{b}\). The midpoint of \(\overline{AC}\) has position vector \(\frac12(\mathbf{a}+\mathbf{c})\). 2. If the diagonals bisect each other, their midpoints are equal: \(\frac12\mathbf{b}=\frac12(\mathbf{a}+\mathbf{c})\). 3. Therefore, \(\mathbf{b}=\mathbf{a}+\mathbf{c}\). 4. This means \(B\) is the endpoint of the sum of the two adjacent side vectors from \(O\). By the parallelogram rule, \(OABC\) is a parallelogram.

1. \(\frac12\mathbf{b}\) and \(\frac12(\mathbf{a}+\mathbf{c})\) 2. \(\mathbf{b}=\mathbf{a}+\mathbf{c}\) 3. \(B\) is obtained by vector addition, so \(OABC\) is a parallelogram.

- A position reached by a translation is found by adding the translation vector to the starting position vector. - The fourth vertex must include both side translations from \(A\). - For part b, subtract the initial endpoint's position vector from the terminal endpoint's position vector. - Compare the two three-vector sums component by component in part c.

1. \(\overrightarrow{OB}=\mathbf{a}+\mathbf{u}=\langle5,1,0\rangle\) and \(\overrightarrow{OD}=\mathbf{a}+\mathbf{v}=\langle2,2,-2\rangle\). 2. The fourth vertex is \(\overrightarrow{OC}=\mathbf{a}+\mathbf{u}+\mathbf{v}=\langle4,3,0\rangle\). 3. \(\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=\langle-1,2,0\rangle=\mathbf{v}\), and \(\overrightarrow{DC}=\overrightarrow{OC}-\overrightarrow{OD}=\langle2,1,2\rangle=\mathbf{u}\). Thus opposite sides have equal direction vectors. 4. Since vector addition is commutative, \(\mathbf{a}+\mathbf{u}+\mathbf{v}=\mathbf{a}+\mathbf{v}+\mathbf{u}=\langle4,3,0\rangle\). The two translations reach the same final point in either order.

a) \(\overrightarrow{OB}=\langle5,1,0\rangle\), \(\overrightarrow{OD}=\langle2,2,-2\rangle\), \(\overrightarrow{OC}=\langle4,3,0\rangle\) b) \(\overrightarrow{BC}=\langle-1,2,0\rangle=\mathbf{v}\) and \(\overrightarrow{DC}=\langle2,1,2\rangle=\mathbf{u}\); opposite sides have matching direction vectors. c) Both sums equal \(\langle4,3,0\rangle\), so the translations may be performed in either order.

- Opposite sides of a parallelogram have the same displacement vector. - From \(B\), use the displacement that takes \(A\) to \(D\). - From \(D\), use the displacement that takes \(A\) to \(B\). - For the final part, subtract position vectors in terminal-minus-initial order.

1. Since \(\overrightarrow{AD}=3\mathbf{v}\), moving from \(B\) by the same vector gives \(\overrightarrow{OC}=\overrightarrow{OB}+3\mathbf{v}=\mathbf{a}+2\mathbf{u}+3\mathbf{v}\). 2. Since \(\overrightarrow{AB}=2\mathbf{u}\), moving from \(D\) by the same vector gives \(\overrightarrow{OC}=\overrightarrow{OD}+2\mathbf{u}=\mathbf{a}+3\mathbf{v}+2\mathbf{u}\). Commutativity of vector addition makes this equal to \(\mathbf{a}+2\mathbf{u}+3\mathbf{v}\). 3. \(\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=3\mathbf{v}\), and \(\overrightarrow{DC}=\overrightarrow{OC}-\overrightarrow{OD}=2\mathbf{u}\).

a) \(\overrightarrow{OC}=\mathbf{a}+2\mathbf{u}+3\mathbf{v}\) b) \(\overrightarrow{OC}=\mathbf{a}+3\mathbf{v}+2\mathbf{u}=\mathbf{a}+2\mathbf{u}+3\mathbf{v}\) c) \(\overrightarrow{BC}=3\mathbf{v}\), \(\overrightarrow{DC}=2\mathbf{u}\)

- Think about the displacement that completes the path from \(P\) to \(Q\) to \(R\). - Read the coordinates of \(Q\) and \(R\) from the diagram. - The missing vector goes from the end of \(\mathbf{u}\) to the end of the resultant.

1. The missing vector must carry the endpoint of \(\mathbf{u}\), \(Q=(4, 0)\), to the endpoint of \(\mathbf{r}\), \(R=(2, 5)\). 2. Thus, \(\mathbf{v}=\overrightarrow{QR}=\langle2-4,5-0\rangle=\langle-2,5\rangle\).

\(\langle-2,5\rangle\)

- Opposite sides of a parallelogram represent equal vectors. - Express the displacement from one point to another as a difference. - Add the vector from \(B\) to \(C\) to point \(A\).

1. In parallelogram \(ABCD\), \(\overrightarrow{AD}=\overrightarrow{BC}\). 2. The vector from \(B\) to \(C\) is \(z_C-z_B=(2+5i)-(4+2i)=-2+3i\). 3. Add this vector to \(z_A\): \(z_D=z_A+(z_C-z_B)=(1+i)+(-2+3i)=-1+4i\).

\(z_D=-1+4i\)

- Use \(\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\). - Rewrite vector subtraction as addition of the opposite vector. - Reorder sums to form head-to-tail chains. - A vector from a point to itself is the zero vector.

1. In part a), replace each subtracted vector by its opposite: \(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}=\overrightarrow{AE}\). 2. In part b), \(\overrightarrow{LM}+\overrightarrow{MN}+\overrightarrow{NL} =\overrightarrow{LN}+\overrightarrow{NL}=\mathbf{0}\). 3. In part c), \(\overrightarrow{XY}+\overrightarrow{ZX}+\overrightarrow{WZ}\). Reorder the sum: \(\overrightarrow{WZ}+\overrightarrow{ZX}+\overrightarrow{XY} =\overrightarrow{WX}+\overrightarrow{XY} =\overrightarrow{WY}\).

a) \(\overrightarrow{AE}\) b) \(\mathbf{0}\) c) \(\overrightarrow{WY}\)

- In a parallelogram, \(\overrightarrow{AB}=\overrightarrow{DC}\). - A rhombus is a parallelogram with four equal sides. - Compare the magnitudes of two adjacent side vectors.

1. \(\overrightarrow{AB}=B-A=\langle 3,0,4\rangle\). 2. Since \(\overrightarrow{DC}=\overrightarrow{AB}\), \(C=D+\overrightarrow{AB}=(1,6,1)+(3,0,4)=(4,6,5)\). 3. \(\|\overrightarrow{AB}\|=\sqrt{3^2+4^2}=5\). 4. \(\overrightarrow{AD}=D-A=\langle 0,5,0\rangle\), so \(\|\overrightarrow{AD}\|=5\). 5. A parallelogram with equal adjacent side lengths is a rhombus. Therefore, \(ABCD\) is a rhombus.

a) \(C=(4,6,5)\) b) \(\|\overrightarrow{AB}\|=\|\overrightarrow{AD}\|=5\), so \(ABCD\) is a rhombus.

- Compute each displacement as terminal point minus initial point. - Add and subtract corresponding components. - Distribute scalar factors to every component. - For part c), follow the directed path from point to point.

1. \(\overrightarrow{PQ}=\langle 2,-4,7\rangle\), \(\overrightarrow{RS}=\langle 2,-4,-3\rangle\), \(\overrightarrow{PR}=\langle -3,-3,6\rangle\), and \(\overrightarrow{QS}=\langle -3,-3,-4\rangle\). 2. \(\mathbf{u}=\langle 2,-4,7\rangle+\langle 2,-4,-3\rangle=\langle 4,-8,4\rangle\). 3. \(\mathbf{v}=2\langle -3,-3,6\rangle-\langle -3,-3,-4\rangle=\langle -3,-3,16\rangle\). 4. \(\overrightarrow{PS}+\overrightarrow{SR}+\overrightarrow{RQ}\) traces a path from \(P\) to \(Q\), so \(\mathbf{w}=\overrightarrow{PQ}=\langle 2,-4,7\rangle\).

a) \(\mathbf{u}=\langle 4,-8,4\rangle\) b) \(\mathbf{v}=\langle -3,-3,16\rangle\) c) \(\mathbf{w}=\langle 2,-4,7\rangle\)

- A vector chain that returns to its starting point has sum \(\mathbf{0}\). - Compute \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) first. - Add \(\overrightarrow{AD}\) to point \(A\) to find \(D\).

1. The vectors in part a) trace the closed path \(A\to B\to C\to A\). Therefore, \(\mathbf{x}=\mathbf{0}\). 2. \(\overrightarrow{AB}=\langle 3,3,-3\rangle\) and \(\overrightarrow{BC}=\langle -6,2,2\rangle\). 3. \(\overrightarrow{AD}=\langle -6,2,2\rangle-2\langle 3,3,-3\rangle=\langle -12,-4,8\rangle\). 4. \(D=A+\overrightarrow{AD}=(2,-1,3)+(-12,-4,8)=(-10,-5,11)\).

a) \(\mathbf{x}=\mathbf{0}\); the vectors form a closed path. b) \(D=(-10,-5,11)\)

- Follow the start and end points in a vector chain. - A zero product can result from either factor being zero. - Compare a direct path with a two-segment path. - Check special cases involving the zero vector.

1. Statement a) is true: a displacement vector is terminal position vector minus initial position vector. 2. Statement b) is true. 3. Statement c) is true because the vectors trace a closed path. 4. Statement d) is false. The equation holds when \(k=0\) or \(\mathbf{a}=\mathbf{0}\). 5. Statement e) is false. In general, \(\lVert\mathbf{a}+\mathbf{b}\rVert\leq\lVert\mathbf{a}\rVert+\lVert\mathbf{b}\rVert\). Equality holds when one vector is a nonnegative scalar multiple of the other, including the case when one vector is \(\mathbf{0}\).

a) True. b) True. c) True. d) False; \(k=0\) or \(\mathbf{a}=\mathbf{0}\). e) False; \(\lVert\mathbf{a}+\mathbf{b}\rVert\leq\lVert\mathbf{a}\rVert+\lVert\mathbf{b}\rVert\).

- Express a vector between two midpoints as a sum of shorter vectors. - Use the fact that a midpoint divides a side vector by \(2\). - Compare a pair of opposite side vectors in \(PQRS\). - Relate both vectors to the same diagonal of \(ABCD\).

1. Since \(P\) and \(Q\) are midpoints, \(\overrightarrow{PQ}=\overrightarrow{PB}+\overrightarrow{BQ} =\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BC} =\frac{1}{2}\overrightarrow{AC}\). 2. Since \(S\) and \(R\) are midpoints, \(\overrightarrow{SR}=\overrightarrow{SD}+\overrightarrow{DR} =\frac{1}{2}\overrightarrow{AD}+\frac{1}{2}\overrightarrow{DC} =\frac{1}{2}\overrightarrow{AC}\). 3. Therefore, \(\overrightarrow{PQ}=\overrightarrow{SR}\). One pair of opposite sides is equal and parallel, so \(PQRS\) is a parallelogram.

\(\overrightarrow{PQ}=\frac{1}{2}\overrightarrow{AC}\) and \(\overrightarrow{SR}=\frac{1}{2}\overrightarrow{AC}\). Hence \(\overrightarrow{PQ}=\overrightarrow{SR}\), so \(PQRS\) is a parallelogram.

- Find each displacement by subtracting coordinates. - Apply scalar multiplication before vector subtraction. - Add and subtract corresponding components. - Simplify the head-to-tail chain in part c).

1. \(\overrightarrow{KL}=\langle -3,7,2\rangle\), \(\overrightarrow{MN}=\langle 5,3,-10\rangle\), \(\overrightarrow{LM}=\langle -3,-5,3\rangle\), and \(\overrightarrow{KN}=\langle -1,5,-5\rangle\). 2. In part a), \(\overrightarrow{KL}+\overrightarrow{MN}=\langle 2,10,-8\rangle\). 3. In part b), \(3\overrightarrow{LM}-\overrightarrow{KN}=\langle -9,-15,9\rangle-\langle -1,5,-5\rangle=\langle -8,-20,14\rangle\). 4. In part c), the vector chain goes from \(K\) to \(N\), so \(\overrightarrow{KL}+\overrightarrow{LM}+\overrightarrow{MN}=\overrightarrow{KN}=\langle -1,5,-5\rangle\).

a) \(\langle 2,10,-8\rangle\) b) \(\langle -8,-20,14\rangle\) c) \(\langle -1,5,-5\rangle\)

- Isolate the position vector of \(D\). - Compute the vectors on the right side first. - Add a known displacement vector to its initial point.

1. \(\overrightarrow{AB}=\langle 4,-4,-2\rangle\) and \(\overrightarrow{BC}=\langle -5,6,5\rangle\). 2. In part a), \(D=C+\overrightarrow{AB}=(0,4,6)+(4,-4,-2)=(4,0,4)\). 3. In part b), \(\overrightarrow{AD}=2\langle 4,-4,-2\rangle-\langle -5,6,5\rangle=\langle 13,-14,-9\rangle\). 4. Therefore, \(D=A+\overrightarrow{AD}=(1,2,3)+(13,-14,-9)=(14,-12,-6)\).

a) \(D=(4,0,4)\) b) \(D=(14,-12,-6)\)

- Find the translation vector as image minus original. - Apply the same vector to every point. - Subtract the translation vector to reverse the transformation.

1. \(\mathbf{v}=Q-P=\langle -2,3,4\rangle\). 2. \(R'=R+\mathbf{v}=(4,1,2)+(-2,3,4)=(2,4,6)\). 3. Reverse the translation: \(S=S'-\mathbf{v}=(-2,10,5)-(-2,3,4)=(0,7,1)\). 4. In general, \((x,y,z)\mapsto(x-2,y+3,z+4)\), so \(X'=(x-2,y+3,z+4)\).

a) \(\mathbf{v}=\langle -2,3,4\rangle\) b) \(R'=(2,4,6)\) c) \(S=(0,7,1)\) d) \(X'=(x-2,y+3,z+4)\)

- A vector to the origin is the opposite of the point's position vector. - Subtract a translation vector to recover the original point. - Returning every point to itself requires zero net displacement.

1. To map \(P\) to the origin, \(\mathbf{a}=\mathbf{0}-P=\langle -2.5,5,-7.2\rangle\). 2. Reverse the translation: \(M=M'-\mathbf{b}=(4.2,1.1,-3.5)-(1.5,-2.4,0.8)=(2.7,3.5,-4.3)\). 3. The net translation must be the zero vector: \(\mathbf{v}_1+\mathbf{v}_2=\mathbf{0}\). Equivalently, \(\mathbf{v}_2=-\mathbf{v}_1\).

a) \(\mathbf{a}=\langle -2.5,5,-7.2\rangle\) b) \(M=(2.7,3.5,-4.3)\) c) \(\mathbf{v}_1+\mathbf{v}_2=\mathbf{0}\), or \(\mathbf{v}_2=-\mathbf{v}_1\)

- Define subtraction as addition of an opposite vector. - Find the opposite of the entire expression in parentheses. - The opposite of an opposite is the original vector. - Work from the innermost parentheses outward.

1. Rewrite the inner subtraction: \(\mathbf{b}-\mathbf{c}=\mathbf{b}+(-\mathbf{c})\). 2. Rewrite the outer subtraction: \(\mathbf{a}-(\mathbf{b}+(-\mathbf{c})) =\mathbf{a}+[-(\mathbf{b}+(-\mathbf{c}))]\). 3. The opposite of a sum is the sum of the opposites: \(-(\mathbf{b}+(-\mathbf{c}))=(-\mathbf{b})+[-(-\mathbf{c})]\). 4. Since \(-(-\mathbf{c})=\mathbf{c}\), \(\mathbf{a}+(-\mathbf{b})+\mathbf{c} =\mathbf{a}-\mathbf{b}+\mathbf{c}\).

\(\mathbf{a}-(\mathbf{b}-\mathbf{c}) =\mathbf{a}+[-\mathbf{b}+\mathbf{c}] =\mathbf{a}-\mathbf{b}+\mathbf{c}\)

- Decompose both vectors through \(M\). - Vectors from a midpoint to the endpoints are opposites. - Combine like vector terms after substitution.

1. Write each vector using \(M\): \(\overrightarrow{AB}=\overrightarrow{AM}+\overrightarrow{MB}\) and \(\overrightarrow{AC}=\overrightarrow{AM}+\overrightarrow{MC}\). 2. Add: \(\overrightarrow{AB}+\overrightarrow{AC} =2\overrightarrow{AM}+\overrightarrow{MB}+\overrightarrow{MC}\). 3. Since \(M\) is the midpoint of \(\overline{BC}\), \(\overrightarrow{MB}=-\overrightarrow{MC}\), so \(\overrightarrow{MB}+\overrightarrow{MC}=\mathbf{0}\). 4. Therefore, \(\overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AM}\).

\(\overrightarrow{AB}+\overrightarrow{AC} =2\overrightarrow{AM}+(\overrightarrow{MB}+\overrightarrow{MC}) =2\overrightarrow{AM}\)

- How are the starting position, displacement, and ending position related? - First combine the two moves into one total displacement. - To reverse a displacement, subtract its vector from the ending position.

1. Add the two displacement vectors: \(\mathbf{w}=\mathbf{u}+\mathbf{v}=\langle 10,-5,2\rangle+\langle 3,8,-4\rangle=\langle 13,3,-2\rangle\). 2. The position vectors satisfy \(\mathbf{p}+\mathbf{w}=\mathbf{q}\), so \(\mathbf{p}=\mathbf{q}-\mathbf{w}\). 3. Therefore, \(\mathbf{p}=\langle 18,12,5\rangle-\langle 13,3,-2\rangle=\langle 5,9,7\rangle\), so \(P(5,9,7)\).

1. \(\mathbf{w}=\langle 13,3,-2\rangle\) 2. \(P(5,9,7)\)

- Every point of a prism's base is translated by the same vector. - The stated correspondence determines the translation vector. - What form must a vector have to be perpendicular to the \(xy\)-plane?

1. In part a), the translation vector is \(\overrightarrow{PS}=\langle 2,2,5\rangle\). Applying this vector gives \(Q'=(7,3,5)\) and \(R'=(3,6,5)\). 2. In part b), the translation vector is \(\overrightarrow{QS}=\langle -2,2,5\rangle\). Applying this vector gives \(P''=(-1,3,5)\) and \(R''=(-1,6,5)\). 3. The base lies in the \(xy\)-plane. A vector perpendicular to that plane has zero \(x\)- and \(y\)-components. Both translation vectors have nonzero horizontal components, so the lateral edges are not perpendicular to the base. Thus, each prism is oblique.

a) \(Q'(7,3,5)\) and \(R'(3,6,5)\) b) \(P''(-1,3,5)\) and \(R''(-1,6,5)\) c) In both cases, the translation vector has nonzero \(x\)- and \(y\)-components, so it is not perpendicular to the \(xy\)-plane. Therefore, the prism is oblique.

- Subtract initial coordinates from terminal coordinates. - Equal vectors have equal corresponding components. - What quadrilateral has one pair of opposite sides equal and parallel? - Translate \(B\) by the vector \(\overrightarrow{AC}\).

1. \(\overrightarrow{AB}=B-A=\langle 3,-2,5\rangle\). 2. \(\overrightarrow{DC}=C-D=\langle 3,-2,5\rangle\). Thus, the vectors are equal. 3. Since one pair of opposite sides is equal and parallel with matching orientation, \(ABCD\) is a parallelogram. 4. \(\overrightarrow{AC}=C-A=\langle -2,3,3\rangle\). 5. Since \(\overrightarrow{BS}=\overrightarrow{AC}\), \(S=B+\overrightarrow{AC}=(3,4,7)\).

a) \(\overrightarrow{AB}=\langle 3,-2,5\rangle\) and \(\overrightarrow{DC}=\langle 3,-2,5\rangle\); they are equal. b) \(ABCD\) is a parallelogram. c) \(S=(3,4,7)\)

- Relate a point, its image, and the translation vector. - To undo a translation, subtract the translation vector. - Consider what happens to the displacement between two points when both are translated by the same vector.

1. Add the translation vector to \(P\): \(P'=(3,1,5)+(-2,4,1)=(1,5,6)\). 2. Reverse the translation to find \(Q\): \(Q=(5,0,2)-(-2,4,1)=(7,-4,1)\). 3. \(\overrightarrow{PQ}=\langle 4,-5,-4\rangle\). 4. \(\overrightarrow{P'Q'}=\langle 4,-5,-4\rangle\). 5. The vectors are equal. Translating both endpoints by the same vector preserves the displacement between them.

a) \(P'(1,5,6)\) b) \(Q(7,-4,1)\) c) \(\overrightarrow{PQ}=\langle 4,-5,-4\rangle\) and \(\overrightarrow{P'Q'}=\langle 4,-5,-4\rangle\). They are equal.

- Which opposite side vectors of a parallelogram are equal? - Use the midpoint formula for each diagonal. - Two segments bisect each other when they have the same midpoint.

1. For consecutive vertices of a parallelogram, \(\overrightarrow{AD}=\overrightarrow{BC}\), so \(D=A+C-B\). 2. Therefore, \(D=(1, 4, -2)+(2, -1, 5)-(5, 2, 2)=(-2, 1, 1)\). 3. The midpoint of \(\overline{AC}\) is \(\left(\frac{1+2}{2},\frac{4+(-1)}{2},\frac{-2+5}{2}\right)=(1.5, 1.5, 1.5)\). 4. The midpoint of \(\overline{BD}\) is \(\left(\frac{5+(-2)}{2},\frac{2+1}{2},\frac{2+1}{2}\right)=(1.5, 1.5, 1.5)\). 5. Since the diagonals have the same midpoint, they bisect each other.

a) \(D(-2, 1, 1)\) b) Both diagonals have midpoint \(M(1.5, 1.5, 1.5)\).

- Opposite sides of a parallelogram can be represented by equal vectors with matching orientation. - Pay attention to the vertex order when choosing the opposite-side vectors. - Equal vectors have equal corresponding components.

1. Compare the opposite-side vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{SR}\): \(\overrightarrow{PQ}=\langle 3-a,4,-2\rangle\), \(\overrightarrow{SR}=\langle -2,4,-2\rangle\). 2. For \(PQRS\) to be a parallelogram, these vectors must be equal. The second and third components already match, and the first components give \(3-a=-2\). 3. Solving gives \(a=5\).

\(a=5\)

- Complete the two reflections one at a time. - A reflection center is the midpoint of a point and its image. - Compare the components of \(\overrightarrow{PP''}\) and \(\overrightarrow{QR}\).

1. For the first point reflection, \(P'=2Q-P=(0,8,7)\). 2. For the second point reflection, \(P''=2R-P'=(6,-12,-5)\). 3. The total displacement is \(\overrightarrow{PP''}=P''-P=\langle 4,-12,-2\rangle\). 4. Also, \(\overrightarrow{QR}=R-Q=\langle 2,-6,-1\rangle\). 5. Therefore, \(2\overrightarrow{QR}=\langle 4,-12,-2\rangle=\overrightarrow{PP''}\).

a) \(P'(0,8,7)\) and \(P''(6,-12,-5)\) b) \(\overrightarrow{PP''}=\langle 4,-12,-2\rangle=2\overrightarrow{QR}\)

- Express each diagonal as a path along edges. - Reverse the sign when moving opposite an edge vector. - Collect the coefficients of \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) separately.

1. Following edge paths gives \(\overrightarrow{AG}=\mathbf{a}+\mathbf{b}+\mathbf{c}\), \(\overrightarrow{BH}=-\mathbf{a}+\mathbf{b}+\mathbf{c}\), \(\overrightarrow{CE}=-\mathbf{a}-\mathbf{b}+\mathbf{c}\), and \(\overrightarrow{DF}=\mathbf{a}-\mathbf{b}+\mathbf{c}\). 2. Add the four expressions. The coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) cancel, while the four \(\mathbf{c}\) terms remain. 3. Therefore, \(\mathbf{s}=4\mathbf{c}\).

a) \(\overrightarrow{AG}=\mathbf{a}+\mathbf{b}+\mathbf{c}\), \(\overrightarrow{BH}=-\mathbf{a}+\mathbf{b}+\mathbf{c}\), \(\overrightarrow{CE}=-\mathbf{a}-\mathbf{b}+\mathbf{c}\), \(\overrightarrow{DF}=\mathbf{a}-\mathbf{b}+\mathbf{c}\) b) \(\mathbf{s}=4\mathbf{c}\)

- Use the correspondence \(A\to D\), \(B\to E\), and \(C\to F\). - Express a base edge such as \(\overrightarrow{BC}\) using \(\mathbf{u}\) and \(\mathbf{v}\). - Collect like vector terms after adding.

1. Using edge paths, \(\overrightarrow{AE}=\mathbf{u}+\mathbf{w}\), \(\overrightarrow{BF}=\mathbf{v}-\mathbf{u}+\mathbf{w}\), and \(\overrightarrow{CD}=-\mathbf{v}+\mathbf{w}\). 2. Adding gives \((\mathbf{u}+\mathbf{w})+(\mathbf{v}-\mathbf{u}+\mathbf{w})+(-\mathbf{v}+\mathbf{w})=3\mathbf{w}\). 3. The other diagonals are \(\overrightarrow{BD}=-\mathbf{u}+\mathbf{w}\), \(\overrightarrow{CE}=\mathbf{u}-\mathbf{v}+\mathbf{w}\), and \(\overrightarrow{AF}=\mathbf{v}+\mathbf{w}\). 4. Their sum is also \(3\mathbf{w}\).

a) \(\overrightarrow{AE}=\mathbf{u}+\mathbf{w}\), \(\overrightarrow{BF}=-\mathbf{u}+\mathbf{v}+\mathbf{w}\), \(\overrightarrow{CD}=-\mathbf{v}+\mathbf{w}\) b) \(3\mathbf{w}\) c) \(\overrightarrow{BD}+\overrightarrow{CE}+\overrightarrow{AF}=3\mathbf{w}\), the same result as in part b).

- Use the equal opposite sides of the parallelogram. - Build each requested vector from known edge vectors. - The diagonals of a parallelogram bisect each other.

1. \(\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=\mathbf{a}+\mathbf{b}\). 2. Opposite sides of a parallelogram are equal as vectors, so \(\overrightarrow{AD}=\mathbf{b}\). 3. \(\overrightarrow{BS}=\overrightarrow{BA}+\overrightarrow{AS}=-\mathbf{a}+\mathbf{s}\). 4. \(\overrightarrow{CS}=\overrightarrow{CB}+\overrightarrow{BA}+\overrightarrow{AS}=-\mathbf{b}-\mathbf{a}+\mathbf{s}\). 5. Since \(M\) is the midpoint of \(\overline{AC}\), \(\overrightarrow{AM}=\frac12(\mathbf{a}+\mathbf{b})\). Thus, \(\overrightarrow{MS}=\mathbf{s}-\frac12\mathbf{a}-\frac12\mathbf{b}\).

a) \(\overrightarrow{AC}=\mathbf{a}+\mathbf{b}\) b) \(\overrightarrow{AD}=\mathbf{b}\) c) \(\overrightarrow{BS}=-\mathbf{a}+\mathbf{s}\) d) \(\overrightarrow{CS}=-\mathbf{a}-\mathbf{b}+\mathbf{s}\) e) \(\overrightarrow{MS}=\mathbf{s}-\frac12\mathbf{a}-\frac12\mathbf{b}\)

- Translate the side relation of the parallelogram directly into a vector equation. - The midpoint of points with position vectors \(\mathbf{p}\) and \(\mathbf{q}\) has position vector \(\frac12(\mathbf{p}+\mathbf{q})\). - Keep point coordinates and displacement vectors distinct when subtracting.

1. \(\overrightarrow{BC}=C-B=\langle-2,4,-2\rangle\), so \(D=A+\overrightarrow{BC}=(0, 3, -2)\). 2. \(A+C=(6, 4, 2)\) and \(B+D=(6, 4, 2)\). Dividing either equal coordinate sum by \(2\) gives the same midpoint for \(\overline{AC}\) and \(\overline{BD}\), so the diagonals bisect each other. 3. \(\overrightarrow{BD}=D-B=\langle-6,2,-6\rangle\).

a) \(D=(0, 3, -2)\) b) \(A+C=B+D=(6, 4, 2)\), so both diagonal midpoints are \((3, 2, 1)\). c) \(\overrightarrow{BD}=\langle-6,2,-6\rangle\)

- Multiply every component by the scalar. - Rearrange the vector equation as you would an algebraic equation. - Check signs carefully when subtracting vectors.

1. \(5\mathbf{a}=\langle10,-15,5\rangle\) and \(2\mathbf{b}=\langle8,2,-4\rangle\). 2. Thus, \(\mathbf{u}=5\mathbf{a}-2\mathbf{b}=\langle2,-17,9\rangle\). 3. For part b, \(2\mathbf{x}=\mathbf{c}-\mathbf{a}-\mathbf{b}\). 4. Since \(\mathbf{a}+\mathbf{b}=\langle6,-2,-1\rangle\), \(2\mathbf{x}=\langle-5,-2,6\rangle\). 5. Therefore, \(\mathbf{x}=\langle-2.5,-1,3\rangle\).

a) \(\mathbf{u}=\langle2,-17,9\rangle\) b) \(\mathbf{x}=\langle-2.5,-1,3\rangle\)

- A displacement vector is terminal point minus initial point. - Work component by component when forming the linear combination. - A point and its position vector have the same coordinates.

1. \(\overrightarrow{PQ}=Q-P=\langle-5,2,5\rangle\) and \(\overrightarrow{QR}=R-Q=\langle1,-5,3\rangle\). 2. \(\mathbf{w}=\langle-5,2,5\rangle+2\langle1,-5,3\rangle=\langle-3,-8,11\rangle\). 3. \(\overrightarrow{PR}=R-P=\langle-4,-3,8\rangle\). 4. \(\overrightarrow{OS}=\langle4,1,-3\rangle+\langle-5,2,5\rangle+\langle-4,-3,8\rangle=\langle-5,0,10\rangle\). 5. Therefore, \(S=(-5, 0, 10)\).

a) \(\overrightarrow{PQ}=\langle-5,2,5\rangle\), \(\overrightarrow{QR}=\langle1,-5,3\rangle\) b) \(\mathbf{w}=\langle-3,-8,11\rangle\) c) \(S=(-5, 0, 10)\)