Free math worksheets

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

Add corresponding components: \(\mathbf{u}+\mathbf{v}=\begin{pmatrix}2+(-5)\\-3+4\end{pmatrix}=\begin{pmatrix}-3\\1\end{pmatrix}\).

\(\begin{pmatrix}-3\\1\end{pmatrix}\)

- Follow the first vector and then the second vector tip to tail. - 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.

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

\(\begin{pmatrix}4\\2\end{pmatrix}\)

- 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: \(\begin{pmatrix}4\\1\\5\end{pmatrix}+\begin{pmatrix}-2\\-7\\10\end{pmatrix} =\begin{pmatrix}2\\-6\\15\end{pmatrix}\). 2. Subtract componentwise: \(\mathbf{v}=\begin{pmatrix}8\\-2\\3\end{pmatrix}-\begin{pmatrix}2\\-6\\15\end{pmatrix} =\begin{pmatrix}6\\4\\-12\end{pmatrix}\). 3. \(\lVert\mathbf{v}\rVert=\sqrt{6^2+4^2+(-12)^2}=\sqrt{196}=14\).

\(\mathbf{v}=\begin{pmatrix}6\\4\\-12\end{pmatrix}\), \(\lVert\mathbf{v}\rVert=14\)

- 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. \(\overrightarrow{BC}=C-B=\begin{pmatrix}-6\\6\\-2\end{pmatrix}\). 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. \(\overrightarrow{PQ}=Q-P=\begin{pmatrix}4\\3\\0\end{pmatrix}\). 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 \(\overrightarrow{HD}=-\vec{c}\), \(\overrightarrow{DA}=-\vec{b}\), and \(\overrightarrow{AB}=\vec{a}\), \(\overrightarrow{HB}=\vec{a}-\vec{b}-\vec{c}\). 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 \(\overrightarrow{EF}=\vec{a}\) and \(\overrightarrow{FG}=\vec{b}\). Therefore, \(\overrightarrow{EG}=\vec{a}+\vec{b}\).

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

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

a) \(\overrightarrow{AB}=\begin{pmatrix}3\\4\\0\end{pmatrix}\), \(\overrightarrow{BA}=\begin{pmatrix}-3\\-4\\0\end{pmatrix}\), and both magnitudes are \(5\). b) \(\overrightarrow{AC}=\begin{pmatrix}3\\0\\-4\end{pmatrix}\), 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. \(\overrightarrow{AB}=B-A=\begin{pmatrix}4\\4\\-3\end{pmatrix}\), and \(\overrightarrow{BC}=C-B=\begin{pmatrix}-7\\3\\6\end{pmatrix}\). 2. \(\mathbf{s}=\begin{pmatrix}4\\4\\-3\end{pmatrix}+\begin{pmatrix}-7\\3\\6\end{pmatrix} =\begin{pmatrix}-3\\7\\3\end{pmatrix}\). 3. \(\overrightarrow{AC}=C-A=\begin{pmatrix}-3\\7\\3\end{pmatrix}\). 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. \(\overrightarrow{AB}=\begin{pmatrix}4\\4\\-3\end{pmatrix}\), \(\overrightarrow{BC}=\begin{pmatrix}-7\\3\\6\end{pmatrix}\) 2. \(\mathbf{s}=\begin{pmatrix}-3\\7\\3\end{pmatrix}\) 3. \(\overrightarrow{AC}=\begin{pmatrix}-3\\7\\3\end{pmatrix}\), 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. \(\overrightarrow{PQ}=Q-P=\begin{pmatrix}4\\6\\-4\end{pmatrix}\). 2. \(-\mathbf{v}=\begin{pmatrix}-4\\-6\\4\end{pmatrix}\). 3. \(\overrightarrow{QP}=P-Q=\begin{pmatrix}-4\\-6\\4\end{pmatrix}\). 4. Therefore, \(-\overrightarrow{PQ}=\overrightarrow{QP}\).

a) \(\overrightarrow{PQ}=\begin{pmatrix}4\\6\\-4\end{pmatrix}\) b) \(-\mathbf{v}=\begin{pmatrix}-4\\-6\\4\end{pmatrix}\) c) \(\overrightarrow{QP}=\begin{pmatrix}-4\\-6\\4\end{pmatrix}\), 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. \(\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}=-\vec{u}+\vec{v}=\vec{v}-\vec{u}\). 2. \(\overrightarrow{CD}=\overrightarrow{CA}+\overrightarrow{AD}=-\vec{v}+\vec{w}=\vec{w}-\vec{v}\). 3. \(\overrightarrow{DB}=\overrightarrow{DA}+\overrightarrow{AB}=-\vec{w}+\vec{u}=\vec{u}-\vec{w}\). 4. When \(\vec{a}\) and \(\vec{b}\) share an initial point, \(\vec{b}-\vec{a}\) is the vector from the terminal point of \(\vec{a}\) to the terminal point of \(\vec{b}\).

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

- 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 \(\mathbf{u}=\mathbf{v}+\mathbf{w} =\begin{pmatrix}-2\\1\\8\end{pmatrix}\).

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

- 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. \(P=(4, 15, 20)-(12, -8, 5)=(-8, 23, 15)\).

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

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

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

- 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 \(A'\): \(\vec{v}=\overrightarrow{AA'}=\begin{pmatrix}2-2\\1-1\\9-1\end{pmatrix}=\begin{pmatrix}0\\0\\8\end{pmatrix}\). 2. Translate \(B\) and \(C\) by \(\vec{v}\): \(B'=(6, 1, 1)+(0, 0, 8)=(6, 1, 9)\) and \(C'=(2, 4, 1)+(0, 0, 8)=(2, 4, 9)\). 3. A lateral edge has length \(\|\vec{v}\|=\sqrt{0^2+0^2+8^2}=8\).

a) \(\vec{v}=\begin{pmatrix}0\\0\\8\end{pmatrix}\) 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. \(\overrightarrow{RS}=S-R=\begin{pmatrix}2-(-1.5)\\-4.5-3\\6-0\end{pmatrix}=\begin{pmatrix}3.5\\-7.5\\6\end{pmatrix}\). 2. \(\overrightarrow{SR}=R-S=\begin{pmatrix}-1.5-2\\3-(-4.5)\\0-6\end{pmatrix}=\begin{pmatrix}-3.5\\7.5\\-6\end{pmatrix}\). 3. Every component is negated, so \(\overrightarrow{SR}=-\overrightarrow{RS}\). The vectors have the same magnitude and opposite directions.

a) \(\overrightarrow{RS}=\begin{pmatrix}3.5\\-7.5\\6\end{pmatrix}\) b) \(\overrightarrow{SR}=\begin{pmatrix}-3.5\\7.5\\-6\end{pmatrix}\) 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: \(\vec{u}=\begin{pmatrix}4+(-3)-1\\-2+5-1\\7+0-(-2)\end{pmatrix}=\begin{pmatrix}0\\2\\9\end{pmatrix}\). 2. For part b, first multiply \(\vec{c}\) by \(3\): \(3\vec{c}=\begin{pmatrix}3\\3\\-6\end{pmatrix}\). 3. Then add \(\vec{a}\): \(\vec{v}=\begin{pmatrix}3+4\\3+(-2)\\-6+7\end{pmatrix}=\begin{pmatrix}7\\1\\1\end{pmatrix}\).

a) \(\vec{u}=\begin{pmatrix}0\\2\\9\end{pmatrix}\) b) \(\vec{v}=\begin{pmatrix}7\\1\\1\end{pmatrix}\)

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

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

1. \(\vec{a}\) 2. Starting at \(B\) and subtracting \(\overrightarrow{AB}\) returns to \(A\). 3. \(\vec{a}-\vec{b}\) points from \(B\) to \(A\), so its opposite is \(\vec{b}-\vec{a}\), 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: \(\begin{pmatrix}8\\-4\\7\end{pmatrix}-\begin{pmatrix}-5\\2\\11\end{pmatrix}=\begin{pmatrix}13\\-6\\-4\end{pmatrix}\). 2. Add the remaining vector: \(\begin{pmatrix}-12\\5\\-3\end{pmatrix}+\begin{pmatrix}13\\-6\\-4\end{pmatrix}=\begin{pmatrix}1\\-1\\-7\end{pmatrix}\).

\(\begin{pmatrix}1\\-1\\-7\end{pmatrix}\)

- 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: \(\begin{pmatrix}4.2\\-1.8\\0\end{pmatrix}-\begin{pmatrix}-0.8\\2.5\\3.7\end{pmatrix}=\begin{pmatrix}5.0\\-4.3\\-3.7\end{pmatrix}\). 2. Evaluate the second pair: \(\begin{pmatrix}5.5\\-6\\1.2\end{pmatrix}+\begin{pmatrix}-2.1\\1.7\\-4.9\end{pmatrix}=\begin{pmatrix}3.4\\-4.3\\-3.7\end{pmatrix}\). 3. Subtract the results: \(\begin{pmatrix}5.0\\-4.3\\-3.7\end{pmatrix}-\begin{pmatrix}3.4\\-4.3\\-3.7\end{pmatrix}=\begin{pmatrix}1.6\\0\\0\end{pmatrix}\).

\(\begin{pmatrix}1.6\\0\\0\end{pmatrix}\)

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

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

- 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 \(\frac12\vec{b}\). The midpoint of \(\overline{AC}\) has position vector \(\frac12(\vec{a}+\vec{c})\). 2. If the diagonals bisect each other, their midpoints are equal: \(\frac12\vec{b}=\frac12(\vec{a}+\vec{c})\). 3. Therefore, \(\vec{b}=\vec{a}+\vec{c}\). 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. \(\frac12\vec{b}\) and \(\frac12(\vec{a}+\vec{c})\) 2. \(\vec{b}=\vec{a}+\vec{c}\) 3. \(B\) is obtained by vector addition, so \(OABC\) is a parallelogram.

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

The missing vector must carry the endpoint of \(\mathbf{u}\) at \(Q=(4, 0)\) to the endpoint of \(\mathbf{r}\) at \(R=(2, 5)\). Thus, \(\mathbf{v}=\overrightarrow{QR}=\begin{pmatrix}2-4\\5-0\end{pmatrix}=\begin{pmatrix}-2\\5\end{pmatrix}\).

\(\begin{pmatrix}-2\\5\end{pmatrix}\)

- 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) \(\overrightarrow{WY}\)

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

1. \(\overrightarrow{AB}=B-A=\begin{pmatrix}5\\2\\-1\end{pmatrix}\) and \(\overrightarrow{AC}=C-A=\begin{pmatrix}3\\2\\2\end{pmatrix}\). 2. Add componentwise: \(\mathbf{s}=\begin{pmatrix}5\\2\\-1\end{pmatrix}+\begin{pmatrix}3\\2\\2\end{pmatrix} =\begin{pmatrix}8\\4\\1\end{pmatrix}\). 3. \(\lVert\mathbf{s}\rVert=\sqrt{8^2+4^2+1^2}=\sqrt{81}=9\).

\(\mathbf{s}=\begin{pmatrix}8\\4\\1\end{pmatrix}\), \(\lVert\mathbf{s}\rVert=9\)

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

a) \(C=(4, 6, 5)\) b) \(\lVert\overrightarrow{AB}\rVert=\lVert\overrightarrow{AD}\rVert=5\), 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. \(\overrightarrow{PQ}=\begin{pmatrix}2\\-4\\7\end{pmatrix}\), \(\overrightarrow{RS}=\begin{pmatrix}2\\-4\\-3\end{pmatrix}\), \(\overrightarrow{PR}=\begin{pmatrix}-3\\-3\\6\end{pmatrix}\), and \(\overrightarrow{QS}=\begin{pmatrix}-3\\-3\\-4\end{pmatrix}\). 2. \(\mathbf{u}=\begin{pmatrix}2\\-4\\7\end{pmatrix}+\begin{pmatrix}2\\-4\\-3\end{pmatrix} =\begin{pmatrix}4\\-8\\4\end{pmatrix}\). 3. \(\mathbf{v}=2\begin{pmatrix}-3\\-3\\6\end{pmatrix}-\begin{pmatrix}-3\\-3\\-4\end{pmatrix} =\begin{pmatrix}-3\\-3\\16\end{pmatrix}\). 4. \(\overrightarrow{PS}+\overrightarrow{SR}+\overrightarrow{RQ}\) traces a path from \(P\) to \(Q\), so \(\mathbf{w}=\overrightarrow{PQ}=\begin{pmatrix}2\\-4\\7\end{pmatrix}\).

a) \(\mathbf{u}=\begin{pmatrix}4\\-8\\4\end{pmatrix}\) b) \(\mathbf{v}=\begin{pmatrix}-3\\-3\\16\end{pmatrix}\) c) \(\mathbf{w}=\begin{pmatrix}2\\-4\\7\end{pmatrix}\)

- 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. \(\overrightarrow{AB}=\begin{pmatrix}3\\3\\-3\end{pmatrix}\) and \(\overrightarrow{BC}=\begin{pmatrix}-6\\2\\2\end{pmatrix}\). 3. \(\overrightarrow{AD}=\begin{pmatrix}-6\\2\\2\end{pmatrix}-2\begin{pmatrix}3\\3\\-3\end{pmatrix} =\begin{pmatrix}-12\\-4\\8\end{pmatrix}\). 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. \(\overrightarrow{KL}=\begin{pmatrix}-3\\7\\2\end{pmatrix}\), \(\overrightarrow{MN}=\begin{pmatrix}5\\3\\-10\end{pmatrix}\), \(\overrightarrow{LM}=\begin{pmatrix}-3\\-5\\3\end{pmatrix}\), and \(\overrightarrow{KN}=\begin{pmatrix}-1\\5\\-5\end{pmatrix}\). 2. In part a), \(\overrightarrow{KL}+\overrightarrow{MN}=\begin{pmatrix}2\\10\\-8\end{pmatrix}\). 3. In part b), \(3\overrightarrow{LM}-\overrightarrow{KN} =\begin{pmatrix}-9\\-15\\9\end{pmatrix}-\begin{pmatrix}-1\\5\\-5\end{pmatrix} =\begin{pmatrix}-8\\-20\\14\end{pmatrix}\). 4. In part c), the vector chain goes from \(K\) to \(N\), so \(\overrightarrow{KL}+\overrightarrow{LM}+\overrightarrow{MN} =\overrightarrow{KN} =\begin{pmatrix}-1\\5\\-5\end{pmatrix}\).

a) \(\begin{pmatrix}2\\10\\-8\end{pmatrix}\) b) \(\begin{pmatrix}-8\\-20\\14\end{pmatrix}\) c) \(\begin{pmatrix}-1\\5\\-5\end{pmatrix}\)

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

1. \(\overrightarrow{AB}=\begin{pmatrix}4\\-4\\-2\end{pmatrix}\) and \(\overrightarrow{BC}=\begin{pmatrix}-5\\6\\5\end{pmatrix}\). 2. In part a), \(D=C+\overrightarrow{AB}=(0, 4, 6)+(4, -4, -2)=(4, 0, 4)\). 3. In part b), \(\overrightarrow{AD} =2\begin{pmatrix}4\\-4\\-2\end{pmatrix}-\begin{pmatrix}-5\\6\\5\end{pmatrix} =\begin{pmatrix}13\\-14\\-9\end{pmatrix}\). 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. \(\mathbf{v}=Q-P=\begin{pmatrix}-2\\3\\4\end{pmatrix}\). 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, \((x, y, z)\mapsto(x-2, y+3, z+4)\).

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

- 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 \(\mathbf{s}=\mathbf{u}+\mathbf{w} =\begin{pmatrix}2\\3\\-4\end{pmatrix}\). 2. Reverse the combined translation: \(A=E-\mathbf{s}\). 3. \(A=(1, 7, -2)-(2, 3, -4)=(-1, 4, 2)\).

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

- 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, \(\mathbf{a}=\mathbf{0}-P=\begin{pmatrix}-2.5\\5\\-7.2\end{pmatrix}\). 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) \(\mathbf{a}=\begin{pmatrix}-2.5\\5\\-7.2\end{pmatrix}\) 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: \(\vec{w}=\vec{u}+\vec{v}=\begin{pmatrix}10\\-5\\2\end{pmatrix}+\begin{pmatrix}3\\8\\-4\end{pmatrix}=\begin{pmatrix}13\\3\\-2\end{pmatrix}\). 2. The position vectors satisfy \(\vec{p}+\vec{w}=\vec{q}\), so \(\vec{p}=\vec{q}-\vec{w}\). 3. Therefore, \(\vec{p}=\begin{pmatrix}18\\12\\5\end{pmatrix}-\begin{pmatrix}13\\3\\-2\end{pmatrix}=\begin{pmatrix}5\\9\\7\end{pmatrix}\), so \(P(5, 9, 7)\).

1. \(\vec{w}=\begin{pmatrix}13\\3\\-2\end{pmatrix}\) 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 \(\overrightarrow{PS}=\begin{pmatrix}3-1\\3-1\\5-0\end{pmatrix}=\begin{pmatrix}2\\2\\5\end{pmatrix}\). 2. Applying this vector gives \(Q'=(5, 1, 0)+(2, 2, 5)=(7, 3, 5)\) and \(R'=(1, 4, 0)+(2, 2, 5)=(3, 6, 5)\). 3. In part b, the translation vector is \(\overrightarrow{QS}=\begin{pmatrix}3-5\\3-1\\5-0\end{pmatrix}=\begin{pmatrix}-2\\2\\5\end{pmatrix}\). 4. Applying this vector gives \(P'=(1, 1, 0)+(-2, 2, 5)=(-1, 3, 5)\) and \(R'= (1, 4, 0)+(-2, 2, 5)=(-1, 6, 5)\). 5. 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. \(\overrightarrow{AB}=B-A=\begin{pmatrix}3\\-2\\5\end{pmatrix}\). 2. \(\overrightarrow{DC}=C-D=\begin{pmatrix}3\\-2\\5\end{pmatrix}\). Thus, the vectors are equal. 3. Since one pair of opposite sides is equal and parallel with matching orientation, \(ABCD\) is a parallelogram. 4. \(\overrightarrow{AC}=C-A=\begin{pmatrix}-2\\3\\3\end{pmatrix}\). 5. Since \(\overrightarrow{BS}=\overrightarrow{AC}\), \(S=B+\overrightarrow{AC}=(5, 1, 4)+(-2, 3, 3)=(3, 4, 7)\).

a) \(\overrightarrow{AB}=\begin{pmatrix}3\\-2\\5\end{pmatrix}\) and \(\overrightarrow{DC}=\begin{pmatrix}3\\-2\\5\end{pmatrix}\); 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. Compute \(\overrightarrow{PQ}=\begin{pmatrix}7-3\\-4-1\\1-5\end{pmatrix}=\begin{pmatrix}4\\-5\\-4\end{pmatrix}\). 4. Compute \(\overrightarrow{P'Q'}=\begin{pmatrix}5-1\\0-5\\2-6\end{pmatrix}=\begin{pmatrix}4\\-5\\-4\end{pmatrix}\). 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) \(\overrightarrow{PQ}=\begin{pmatrix}4\\-5\\-4\end{pmatrix}\) and \(\overrightarrow{P'Q'}=\begin{pmatrix}4\\-5\\-4\end{pmatrix}\). 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 \(\overrightarrow{SR}\): \(\overrightarrow{PQ}=Q-P=\begin{pmatrix}3-a\\4\\-2\end{pmatrix}\), \(\overrightarrow{SR}=R-S=\begin{pmatrix}-2\\4\\-2\end{pmatrix}\). 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, \(P'=2Q-P=2(1, 4, 2)-(2, 0, -3)=(0, 8, 7)\). 2. For the second point reflection, \(P''=2R-P'=2(3, -2, 1)-(0, 8, 7)=(6, -12, -5)\). 3. The total displacement is \(\overrightarrow{PP''}=P''-P=(6, -12, -5)-(2, 0, -3)=(4, -12, -2)\). 4. Also, \(\overrightarrow{QR}=R-Q=(3, -2, 1)-(1, 4, 2)=(2, -6, -1)\). 5. Therefore, \(2\overrightarrow{QR}=(4, -12, -2)=\overrightarrow{PP''}\).

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

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

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

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

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

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

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

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

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

\(\vec{x}=\begin{pmatrix}-18\\8\\2\end{pmatrix}\)

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

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

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

1. \(5\vec{a}=\begin{pmatrix}10\\-15\\5\end{pmatrix}\) and \(2\vec{b}=\begin{pmatrix}8\\2\\-4\end{pmatrix}\). 2. Thus, \(\vec{u}=5\vec{a}-2\vec{b}=\begin{pmatrix}2\\-17\\9\end{pmatrix}\). 3. For part b, \(2\vec{x}=\vec{c}-\vec{a}-\vec{b}\). 4. Since \(\vec{a}+\vec{b}=\begin{pmatrix}6\\-2\\-1\end{pmatrix}\), \(2\vec{x}=\begin{pmatrix}-5\\-2\\6\end{pmatrix}\). 5. Therefore, \(\vec{x}=\begin{pmatrix}-2.5\\-1\\3\end{pmatrix}\).

a) \(\vec{u}=\begin{pmatrix}2\\-17\\9\end{pmatrix}\) b) \(\vec{x}=\begin{pmatrix}-2.5\\-1\\3\end{pmatrix}\)

- 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. \(\overrightarrow{PQ}=Q-P=\begin{pmatrix}-5\\2\\5\end{pmatrix}\) and \(\overrightarrow{QR}=R-Q=\begin{pmatrix}1\\-5\\3\end{pmatrix}\). 2. \(\vec{w}=\begin{pmatrix}-5\\2\\5\end{pmatrix}+2\begin{pmatrix}1\\-5\\3\end{pmatrix}=\begin{pmatrix}-3\\-8\\11\end{pmatrix}\). 3. \(\overrightarrow{PR}=R-P=\begin{pmatrix}-4\\-3\\8\end{pmatrix}\). 4. \(\overrightarrow{OS}=\begin{pmatrix}4\\1\\-3\end{pmatrix}+\begin{pmatrix}-5\\2\\5\end{pmatrix}+\begin{pmatrix}-4\\-3\\8\end{pmatrix}=\begin{pmatrix}-5\\0\\10\end{pmatrix}\). 5. Therefore, \(S(-5, 0, 10)\).

a) \(\overrightarrow{PQ}=\begin{pmatrix}-5\\2\\5\end{pmatrix}\), \(\overrightarrow{QR}=\begin{pmatrix}1\\-5\\3\end{pmatrix}\) b) \(\vec{w}=\begin{pmatrix}-3\\-8\\11\end{pmatrix}\) c) \(S(-5, 0, 10)\)