Free Math Worksheets

- 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. \(\lVert\mathbf{v}\rVert=\sqrt{6^2+4^2+(-12)^2}=\sqrt{196}=14\).

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

- 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 \(\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\).

- 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) \(\mathbf{u}=\begin{pmatrix}3\\-8\\5\end{pmatrix}=-\mathbf{v}\)

- 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.

- 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. \(\lVert\mathbf{s}\rVert=\sqrt{8^2+4^2+1^2}=\sqrt{81}=9\).

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

- 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, \(S=R-\overrightarrow{PQ}=(5, 6, 0)-(4, 3, 0)=(1, 3, 0)\).

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

- 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) \(D=(14, -12, -6)\)

- 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 \(A\).

- 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. \(P(5, 9, 7)\)

- 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.

- 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) \(S(-5, 0, 10)\)