Exercises

Exercises 2 Exercises

Vectors associated to directed line segments.

In each case, compute the components of the vector $\vec{v}$ that corresponds to the directed line segment $\abray{PQ}\text{.}$ Then draw both that directed line segment, and the vector $\vec{v}$ with its initial point at the origin.

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1.

$P(3,5) \text{,}$ $Q(7,4) $

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Answer.

$\vec{v} = (7 - 3, 4 - 5) = (4,-1) $

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Diagram illustrating vectors as representing directed line segments.

A diagram consisting of a directed line segment, along with the corresponding vector “based” at the origin. The directed line segment $\abray{PQ}$ extends from the point $P(3,5)$ to the point $Q(7,4)\text{.}$ The corresponding vector $\vec{v}$ is drawn as a directed line segment from the origin to the point $R(4,-1)\text{.}$

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2.

$P(2,-1) \text{,}$ $Q(-1,2) $

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Answer.

$\vec{v} = \bbrac{-1 - 2, 2 - (-1)} = (-3,3) $

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A diagram consisting of a directed line segment, along with the corresponding vector “based” at the origin. The directed line segment $\abray{PQ}$ extends from the point $P(2,-1)$ to the point $Q(-1,2)\text{.}$ The corresponding vector $\vec{v}$ is drawn as a directed line segment from the origin to the point $R(-3,3)\text{.}$

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3.

$P(3,6) \text{,}$ $Q(-3,3) $

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Answer.

$\vec{v} = (-3 - 3, 3 - 6) = (-6,-3) $

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A diagram consisting of a directed line segment, along with the corresponding vector “based” at the origin. The directed line segment $\abray{PQ}$ extends from the point $P(3,6)$ to the point $Q(-3,3)\text{.}$ The corresponding vector $\vec{v}$ is drawn as a directed line segment from the origin to the point $R(-6,-3)\text{.}$

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4.

$P\left( - \dfrac{6}{5}, - \dfrac{ 2}{9} \right) \text{,}$ $Q\left( \dfrac{3}{5}, \dfrac{11}{9} \right) $

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Answer.

$\vec{v} = \left( \dfrac{ 3}{5} - \left(- \dfrac{6}{5} \right), \dfrac{11}{9} - \left(- \dfrac{2}{9} \right) \right) = \left( \dfrac{9}{5}, \dfrac{13}{9} \right)$

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A diagram consisting of a directed line segment, along with the corresponding vector “based” at the origin. The directed line segment $\abray{PQ}$ extends from the point $P\left(-\frac{6}{5},-\frac{2}{9}\right)$ to the point $Q\left(\frac{3}{5},\frac{11}{9}\right)\text{.}$ The corresponding vector $\vec{v}$ is drawn as a directed line segment from the origin to the point $R(-6,-3)\text{.}$

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5.

We can use vectors to represent other kinds of “displacements” besides position displacements. Vectors can be used to represent change between states of any collection of related variables.

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For example, suppose an investor sinks $10,000 into stock in each of companies A, B, C, and D. After a year, the various items in their portfolio have the following values: company A, $10,475; company B, $11,240; company C, $9,756; company D, $10,054.

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Represent the “displacement” in the collection of values of the investor’s holdings, from initial state

\begin{equation*} (A,B,C,D) = (10000, 10000, 10000, 10000) \end{equation*}

to terminal state

\begin{equation*} (A,B,C,D) = (10475, 11240, 9756, 10054) \end{equation*}

as a four-dimensional “profit” vector $\vec{p} \text{.}$

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Answer.

$\vec{p} = (475, 1240, -244, 54) $

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Recognizing the dimension of a vector.

In each case, determine the value of $n $ for which the given vector is a member of the collection $\R^n \text{.}$

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6.

$\vec{a} = (4,0,6) $

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Answer.

$n = 3 $

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7.

$\vec{b} = (-2,4) $

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Answer.

$n = 2 $

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8.

$\vec{c} = (\pi,\sqrt{2},-\pi,e) $

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Answer.

$n = 4 $

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Vector equality.

In each case, determine the value(s) of $k \text{,}$ if any, for which the vector equality is true.

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9.

$(k + 2, k^2, 1, k + 6) = (-3, 25, 1, 1) $

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Answer.

$k = -5 $

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10.

$(k + 2, k^2, 1, k + 6) = (-3, 25, -1, 1) $

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Answer.

There is no such value of $k \text{,}$ as the two vectors will always disagree in the third component.

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11.

$(k + 2, k^2, 1, k + 6, 0) = (-3, 25, 1, 1) $

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Answer.

There is no such value of $k \text{,}$ as the first vector is $5$-dimensional while the second vector is only $4$-dimensional.

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Geometric addition of vectors.

In each case, draw a vector addition triangle in the Cartesian plane that represents $\vec{w} = \vec{u} + \vec{v}\text{,}$ with the initial point of $\vec{u}$ at the origin. Then calculate the sum vector $\vec{w}$ and verify that its terminal point (from the origin) agrees with the terminal point of addition chain $\vec{u} + \vec{v}$ on your diagram.

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12.

$\vec{u} = (-2,3) \text{,}$ $\vec{v} = (6,0) $

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Answer.

A diagram consisting of a triangle in the $xy$-plane, with a directed line segment for each side. One side is a vector labelled $\vec{u}\text{,}$ extending from the origin to the point $P(-2,3)\text{.}$ The next side is a vector labelled $\vec{v}\text{,}$ extending from $P$ to the point $Q(4,3)\text{.}$ The last side is a vector labelled $\vec{w}\text{,}$ extending from the origin to $Q\text{.}$

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$\vec{w} = (-2,3) + (6,0) = (4,3) $

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13.

$\vec{u} = (6,0) \text{,}$ $\vec{v} = (-2,3) $

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Answer.

A diagram consisting of a triangle in the $xy$-plane, with a directed line segment for each side. One side is a vector labelled $\vec{u}\text{,}$ extending from the origin to the point $P(6,0)\text{.}$ The next side is a vector labelled $\vec{v}\text{,}$ extending from $P$ to the point $Q(4,3)\text{.}$ The last side is a vector labelled $\vec{w}\text{,}$ extending from the origin to $Q\text{.}$

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$\vec{w} = (6,0) + (-2,3) = (4,3) $

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14.

$\vec{u} = (-2,-4) \text{,}$ $\vec{v} = (0,3) $

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Answer.

A diagram consisting of a triangle in the $xy$-plane, with a directed line segment for each side. One side is a vector labelled $\vec{u}\text{,}$ extending from the origin to the point $P(-2,-4)\text{.}$ The next side is a vector labelled $\vec{v}\text{,}$ extending from $P$ to the point $Q(-2,-1)\text{.}$ The last side is a vector labelled $\vec{w}\text{,}$ extending from the origin to $Q\text{.}$

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$\vec{w} = (-2,-4) + (0,3) = (-2,-1) $

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15.

$\vec{u} = (0,3) \text{,}$ $\vec{v} = (-2,-4) $

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Answer.

A diagram consisting of a triangle in the $xy$-plane, with a directed line segment for each side. One side is a vector labelled $\vec{u}\text{,}$ extending from the origin to the point $P(0,3)\text{.}$ The next side is a vector labelled $\vec{v}\text{,}$ extending from $P$ to the point $Q(-2,-1)\text{.}$ The last side is a vector labelled $\vec{w}\text{,}$ extending from the origin to $Q\text{.}$

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$\vec{w} = (0,3) + (-2,-4) = (-2,-1) $

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16.

$\vec{u} = (6,4) \text{,}$ $\vec{v} = (-2,-5) $

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Answer.

A diagram consisting of a triangle in the $xy$-plane, with a directed line segment for each side. One side is a vector labelled $\vec{u}\text{,}$ extending from the origin to the point $P(6,4)\text{.}$ The next side is a vector labelled $\vec{v}\text{,}$ extending from $P$ to the point $Q(4,-1)\text{.}$ The last side is a vector labelled $\vec{w}\text{,}$ extending from the origin to $Q\text{.}$

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$\vec{w} = (6,4) + (-2,-5) = (4,-1) $

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17.

$\vec{u} = (-2,-5) \text{,}$ $\vec{v} = (6,4) $

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Answer.

A diagram consisting of a triangle in the $xy$-plane, with a directed line segment for each side. One side is a vector labelled $\vec{u}\text{,}$ extending from the origin to the point $P(-2,-5)\text{.}$ The next side is a vector labelled $\vec{v}\text{,}$ extending from $P$ to the point $Q(4,-1)\text{.}$ The last side is a vector labelled $\vec{w}\text{,}$ extending from the origin to $Q\text{.}$

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$\vec{w} = (-2,-5) + (6,4) = (4,-1) $

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Geometric negation of vectors.

In each case, draw both the vector $\vec{v}$ and its negative in the Cartesian plane with initial points at the origin. Then calculate the negative vector $- \vec{v}$ and verify that its terminal point on your diagram agrees with your calculation.

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18.

$\vec{v} = (5,0) $

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Answer.

A diagram consisting of two oppositely directed line segments in the $xy$-plane. One represents a vector labelled $\vec{v}\text{,}$ and extends from the origin to the point $P(5,0)\text{.}$ The other represents the negative vector $- \vec{v}\text{,}$ and extends from the origin to the point $P'(-5,0)\text{.}$

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$- \vec{v} = - (5,0) = (-5,0) $

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19.

$\vec{v} = (0,-3) $

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Answer.

A diagram consisting of two oppositely directed line segments in the $xy$-plane. One represents a vector labelled $\vec{v}\text{,}$ and extends from the origin to the point $P(0,-3)\text{.}$ The other represents the negative vector $- \vec{v}\text{,}$ and extends from the origin to the point $P'(0,3)\text{.}$

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$- \vec{v} = - (0,-3) = (0,3) $

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20.

$\vec{v} = (6,1) $

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Answer.

A diagram consisting of two oppositely directed line segments in the $xy$-plane. One represents a vector labelled $\vec{v}\text{,}$ and extends from the origin to the point $P(6,1)\text{.}$ The other represents the negative vector $- \vec{v}\text{,}$ and extends from the origin to the point $P'(-6,-1)\text{.}$

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$- \vec{v} = - (6,1) = (-6,-1) $

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21.

$\vec{v} = (-3,2) $

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Answer.

A diagram consisting of two oppositely directed line segments in the $xy$-plane. One represents a vector labelled $\vec{v}\text{,}$ and extends from the origin to the point $P(-3,2)\text{.}$ The other represents the negative vector $- \vec{v}\text{,}$ and extends from the origin to the point $P'(3,-2)\text{.}$

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$- \vec{v} = - (-3,2) = (3,-2) $

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Geometric scaling of vectors.

In each case, draw both the vector $\vec{v}$ and the scaled vector $k \vec{v}$ in the Cartesian plane with initial points at the origin. Then calculate $k \vec{v}$ and verify that its terminal point on your diagram agrees with your calculation.

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22.

$\displaystyle \vec{v} = (-2,0) \text{,}$ $k = 5$

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Answer.

Diagram illustrating scaling of a vector.

A diagram consisting of two parallel directed line segments in the $xy$-plane. One represents a vector labelled $\vec{v}\text{,}$ and extends from the origin to the point $P(-2,0)\text{.}$ The other represents the scaled vector $5 \vec{v}\text{,}$ and extends from the origin to the point $P'(-10,0)\text{.}$

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$5 \vec{v} = 5 (2,0) = (10,0) $

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23.

$\displaystyle \vec{v} = \left(0, \frac{5}{4}\right) \text{,}$ $k = 3$

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Answer.

A diagram consisting of two parallel directed line segments in the $xy$-plane. One represents a vector labelled $\vec{v}\text{,}$ and extends from the origin to the point $P\left(0,\frac{5}{4}\right)\text{.}$ The other represents the scaled vector $5 \vec{v}\text{,}$ and extends from the origin to the point $P'\left(0,\frac{15}{4}\right)\text{.}$

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$3 \vec{v} = 3 \left(0,\frac{5}{4}\right) = \left(0,\frac{15}{4}\right) $

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24.

$\displaystyle \vec{v} = (2, 6) \text{,}$ $k = \dfrac{2}{3}$

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Answer.

A diagram consisting of two parallel directed line segments in the $xy$-plane. One represents a vector labelled $\vec{v}\text{,}$ and extends from the origin to the point $P(2,6)\text{.}$ The other represents the scaled vector $\frac{2}{3} \vec{v}\text{,}$ and extends from the origin to the point $P'\left(\frac{4}{3},4\right)\text{.}$

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$\displaystyle \frac{2}{3} \vec{v} = \frac{2}{3} (2,6) = \left(\frac{4}{3},4\right) $

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25.

$\displaystyle \vec{v} = (-2, 3) \text{,}$ $k = -\dfrac{3}{2}$

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Answer.

A diagram consisting of two parallel directed line segments in the $xy$-plane. One represents a vector labelled $\vec{v}\text{,}$ and extends from the origin to the point $P(-2,3)\text{.}$ The other represents the scaled vector $- \frac{3}{2} \vec{v}\text{,}$ and extends from the origin to the point $P'\left(3,-\frac{9}{2}\right)\text{.}$

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$\displaystyle - \frac{3}{2} \vec{v} = - \frac{3}{2} (-2,3) = \left(3, - \frac{9}{3}\right) $

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Geometric subtraction of vectors.

In each case, draw a vector difference triangle in the Cartesian plane that represents $\vec{w} = \vec{v} - \vec{u}\text{,}$ with the initial points of both $\vec{u}$ and $\vec{v}$ at the origin. Then calculate the difference vector $\vec{w}$ and verify that its components represent the corresponding displacement on your diagram.

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26.

$\vec{u} = (6,0) \text{,}$ $\vec{v} = (-2,3) $

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Answer.

Diagram illustrating vector subtraction.

A diagram consisting of a triangle in the $xy$-plane, with a directed line segment for each side. One side is a vector labelled $\vec{u}\text{,}$ extending from the origin to the point $P(6,0)\text{.}$ The next side is a vector labelled $\vec{v}\text{,}$ extending from the origin to the point $Q(-2,3)\text{.}$ The last side is a vector labelled $\vec{w}\text{,}$ extending from $P$ to $Q\text{.}$

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$\vec{w} = (-2,3) - (6,0) = (-8,3) $

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27.

$\vec{u} = (-2,3) \text{,}$ $\vec{v} = (6,0) $

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Answer.

A diagram consisting of a triangle in the $xy$-plane, with a directed line segment for each side. One side is a vector labelled $\vec{u}\text{,}$ extending from the origin to the point $P(-2,3)\text{.}$ Another side is a vector labelled $\vec{v}\text{,}$ extending from the origin to the point $Q(6,0)\text{.}$ The last side is a vector labelled $\vec{w}\text{,}$ extending from $P$ to $Q\text{.}$

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$\vec{w} = (6,0) - (-2,3) = (8,-3) $

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28.

$\vec{u} = (0,3) \text{,}$ $\vec{v} = (-2,-4) $

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Answer.

A diagram consisting of a triangle in the $xy$-plane, with a directed line segment for each side. One side is a vector labelled $\vec{u}\text{,}$ extending from the origin to the point $P(0,3)\text{.}$ The next side is a vector labelled $\vec{v}\text{,}$ extending from the origin to the point $Q(-2,-4)\text{.}$ The last side is a vector labelled $\vec{w}\text{,}$ extending from $P$ to $Q\text{.}$

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$\vec{w} = (-2,-4) - (0,3) = (-2,-7) $

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29.

$\vec{u} = (-2,-4) \text{,}$ $\vec{v} = (0,3) $

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Answer.

A diagram consisting of a triangle in the $xy$-plane, with a directed line segment for each side. One side is a vector labelled $\vec{u}\text{,}$ extending from the origin to the point $P(-2,-4)\text{.}$ The next side is a vector labelled $\vec{v}\text{,}$ extending from the origin to the point $Q(0,3)\text{.}$ The last side is a vector labelled $\vec{w}\text{,}$ extending from $P$ to $Q\text{.}$

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$\vec{w} = (0,3) - (-2,-4) = (2,7) $

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30.

$\vec{u} = (-2,-5) \text{,}$ $\vec{v} = (6,4) $

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Answer.

A diagram consisting of a triangle in the $xy$-plane, with a directed line segment for each side. One side is a vector labelled $\vec{u}\text{,}$ extending from the origin to the point $P(-2,-5)\text{.}$ The next side is a vector labelled $\vec{v}\text{,}$ extending from the origin to the point $Q(6,4)\text{.}$ The last side is a vector labelled $\vec{w}\text{,}$ extending from $P$ to $Q\text{.}$

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$\vec{w} = (6,4) - (-2,-5) = (8,9) $

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31.

$\vec{u} = (6,4) \text{,}$ $\vec{v} = (-2,-5) $

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Answer.

A diagram consisting of a triangle in the $xy$-plane, with a directed line segment for each side. One side is a vector labelled $\vec{u}\text{,}$ extending from the origin to the point $P(6,4)\text{.}$ The next side is a vector labelled $\vec{v}\text{,}$ extending from $P$ to the point $Q(-2,-5)\text{.}$ The last side is a vector labelled $\vec{w}\text{,}$ extending from $P$ to $Q\text{.}$

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$\vec{w} = (-2,-5) - (6,4) = (-8,-9) $

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Geometric relationships.

In each case, state whether the equality of vectors is true or false based on the diagram below.

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Diagram of several vectors in various geometric relationships.

A planar diagram consisting of six points, $P,Q,R,S,T,U \text{.}$ Quadrilateral $\abcdquad{PQTS} $ is a parallelogram. Side $\absegment{PQ} $ is extended past $Q $ to a point $R$ and side $\absegment{ST} $ is extended past $T $ to a point $U $ so that quadrilateral $\abcdquad{QRUS} $ is also a parallelogram, with $T $ at the midpoint of side $\absegment{SU} \text{.}$ Finally, the following directed line segments are drawn in:

$\abray{PQ} \text{,}$ labelled as vector $\vec{a} \text{.}$

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$\abray{PS} \text{,}$ labelled as vector $\vec{b} \text{.}$

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$\abray{SQ} \text{,}$ labelled as vector $\vec{c} \text{.}$

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$\abray{TS} \text{,}$ labelled as vector $\vec{d} \text{.}$

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$\abray{QT} \text{,}$ labelled as vector $\vec{e} \text{.}$

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$\abray{TU} \text{,}$ labelled as vector $\vec{f} \text{.}$

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$\abray{QR} \text{,}$ labelled as vector $\vec{g} \text{.}$

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$\abray{RT} \text{,}$ labelled as vector $\vec{h} \text{.}$

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$\abray{UR} \text{,}$ labelled as vector $\vec{i} \text{.}$

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32.

$\vec{f} = \vec{a} $

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Answer.

True.

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33.

$- \vec{d} = \vec{a} $

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Answer.

True.

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34.

$\vec{c} + \vec{i} = \zerovec $

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Answer.

True.

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35.

$\vec{b} + \vec{c} = \vec{a} $

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Answer.

True.

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36.

$\vec{d} + \vec{e} = \vec{i} $

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Answer.

True.

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37.

$\vec{e} - \vec{g} = \vec{h} $

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Answer.

True.

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38.

$\vec{i} - \vec{h} = \vec{f} $

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Answer.

True.

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39.

$\vec{f} - \vec{d} = \vec{g} $

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Answer.

True.

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40.

$2 \vec{a} = \vec{g} $

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Answer.

True.

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41.

$\vec{e} + \vec{f} - \vec{i} = \vec{g} $

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Answer.

True.

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42.

$\vec{b} - \vec{d} - \vec{e} = \vec{a} $

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Answer.

True.

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43.

$\vec{c} + \vec{g} + \vec{i} = 2 \vec{a} $

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Answer.

True.

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Recognizing parallel vectors.

In each case, determine whether the two vectors are parallel.

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44.

$\vec{u} = (-3,4) \text{,}$ $\vec{v} = (-12,16) $

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Answer.

Yes, $\vec{v} = 4 \vec{u} \text{.}$

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45.

$\vec{u} = (5,-8) \text{,}$ $\vec{v} = (10,-17) $

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Answer.

No.

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46.

$\vec{u} = (18,-6) \text{,}$ $\vec{v} = (-12,4) $

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Answer.

Yes, $\vec{u} = -\frac{3}{2} \vec{v} \text{.}$

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47.

$\vec{u} = (-6,3,-1) \text{,}$ $\vec{v} = (-2,1,-1/3) $

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Answer.

Yes, $\vec{u} = 3 \vec{v} \text{.}$

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48.

$\vec{u} = (20,-5,10) \text{,}$ $\vec{v} = (32,-8,16) $

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Answer.

Yes, $\vec{v} = \frac{8}{5} \vec{u} \text{.}$

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49.

$\vec{u} = (7,2,-9) \text{,}$ $\vec{v} = (14,6,-1) $

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Answer.

No.

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Vector arithmetic.

In each case, carry out the vector operation(s).

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50.

$(-1,0) + (0,5) $

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Answer.

$(-1,5)$

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51.

$(7,3) + (2,7) $

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Answer.

$(9,10)$

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52.

$(-3,8,1) + (7,8,-3) $

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Answer.

$(4,16,-2)$

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53.

$(-3,0,0) + (0,8,0) + (0,0,3) $

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Answer.

$(-3,8,3)$

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54.

$(-8,1,0,5) + (-3,0,-1,5) $

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Answer.

$(-11,1,-1,10)$

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55.

$(-3,0,0,0) + (0,8,0,0), + (0,0,-2,0) + (0,0,0,5) $

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Answer.

$(-3,8,-2,5)$

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56.

$(1,-5,9,0,3) + (2,4,-7,-8,1) $

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Answer.

$(3,-1,2,-8,4)$

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57.

$(8,0,0,0,0) + (0,4,0,0,0) + (0,0,-9,0,0) + (0,0,0,-7,0) + (0,0,0,0,7) $

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Answer.

$(8,4,-9,-7,7)$

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58.

$(5,2) - (1,3) $

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Answer.

$(4,-1)$

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59.

$(7,-9,-6) - (7,3,-2) $

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Answer.

$(0,-12,-4)$

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60.

$(-4,1,-2,6) - (8,9,-2,-4) $

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Answer.

$(-12,-8,0,10)$

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61.

$(8,9,7,2,-3) - (-9,2,0,4,8) $

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Answer.

$(17,7,7,-2,-11)$

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62.

$-7 (1,3) $

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Answer.

$(-7,-21)$

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63.

$\dfrac{6}{5} (-9,4) $

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Answer.

$\left( - \dfrac{54}{5}, \frac{24}{5} \right)$

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64.

$3 (5,6,1) $

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Answer.

$(15,18,3)$

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65.

$7 \left(1,2,-4, \dfrac{8}{7} \right) $

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Answer.

$(7,14,-28,8)$

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66.

$2 (1,0) + (-3) (0,1) $

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Answer.

$(2,-3)$

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67.

$6 (-1,2) + 7 (-2,-3) $

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Answer.

$(-20,-9)$

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68.

$8 (-3,1,5) + 7 (-1,0,-3) $

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Answer.

$(-31,8,19)$

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69.

$(-5) (2,-4,9,-7) + 5 (-8,-7,-3,-7) + 7 (3,5,-5,-8) $

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Answer.

$(-29,20,-95,-56)$

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Decomposing vectors in $\R^2$ in terms of standard basis vectors.

In each case, express the vector $\vec{v}$ as a linear combination of the standard basis vectors $\vec{e}_1$ and $\vec{e}_2\text{.}$ Then draw a vector addition triangle in the Cartesian plane that represents your decomposition.

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70.

$\vec{v} = (3,4) $

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Answer.

$(3,4) = 3 (1,0) + 4 (0,1) $

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Diagram illustrating a decomposition relative to the standard basis vectors in two dimensions.

A diagram consisting of a right triangle in the $xy$-plane, with a directed line segment for each side. One leg of the triangle is a vector lying along the horizontal axis, extending from the origin to the point $P(3,0)\text{.}$ This vector is labelled $3 \vec{e}_1\text{.}$ The other leg of the triangle is a vector running parallel to the vertical axis, extending from $P$ to the point $Q(3,4)\text{.}$ This vector is labelled $4 \vec{e}_2\text{.}$ The hypotenuse of the triangle is a vector labelled $\vec{v}\text{,}$ extending from the origin to $Q\text{.}$

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71.

$\vec{v} = (5,-2) $

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Answer.

$(5,-2) = 5 (1,0) + (-2) (0,1) $

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A diagram consisting of a right triangle in the $xy$-plane, with a directed line segment for each side. One leg of the triangle is a vector lying along the horizontal axis, extending from the origin to the point $P(5,0)\text{.}$ This vector is labelled $3 \vec{e}_1\text{.}$ The other leg of the triangle is a vector running parallel to the vertical axis, extending from $P$ to the point $Q(5,-2)\text{.}$ This vector is labelled $-2 \vec{e}_2\text{.}$ The hypotenuse of the triangle is a vector labelled $\vec{v}\text{,}$ extending from the origin to $Q\text{.}$

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72.

$\vec{v} = (-3,1) $

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Answer.

$(-3,1) = (-3) (1,0) + 1 (0,1) $

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A diagram consisting of a right triangle in the $xy$-plane, with a directed line segment for each side. One leg of the triangle is a vector lying along the horizontal axis, extending from the origin to the point $P(-3,0)\text{.}$ This vector is labelled $-3 \vec{e}_1\text{.}$ The other leg of the triangle is a vector running parallel to the vertical axis, extending from $P$ to the point $Q(-3,1)\text{.}$ This vector is labelled $\vec{e}_2\text{.}$ The hypotenuse of the triangle is a vector labelled $\vec{v}\text{,}$ extending from the origin to $Q\text{.}$

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73.

$\vec{v} = \left(- \dfrac{3}{2}, - \dfrac{7}{8} \right) $

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Answer.

$\left( - \dfrac{3}{2}, - \dfrac{7}{8} \right) = \left(- \frac{3}{2}\right) (1,0) + \left(- \frac{7}{8}\right) (0,1) $

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A diagram consisting of a right triangle in the $xy$-plane, with a directed line segment for each side. One leg of the triangle is a vector lying along the horizontal axis, extending from the origin to the point $P(-\frac{3}{2},0)\text{.}$ This vector is labelled $-\frac{3}{2} \vec{e}_1\text{.}$ The other leg of the triangle is a vector running parallel to the vertical axis, extending from $P$ to the point $Q(-\frac{3}{2},-\frac{7}{8})\text{.}$ This vector is labelled $-\frac{7}{8} \vec{e}_2\text{.}$ The hypotenuse of the triangle is a vector labelled $\vec{v}\text{,}$ extending from the origin to $Q\text{.}$

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Decomposing vectors in $\R^n$ in terms of standard basis vectors.

In each case, express the vector $\vec{v}$ as a linear combination of the standard basis vectors $\vec{e}_1,\dotsc,\vec{e}_n\text{.}$

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74.

$\vec{v} = (6,5,2) $

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Answer.

$(6,5,2) = 6 \vec{e}_1 + 5 \vec{e}_2 + 2 \vec{e}_3 $

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75.

$\vec{v} = (0,-3,6) $

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Answer.

$(0,-3,6) = 0 \vec{e}_1 + (-3) \vec{e}_2 + 6 \vec{e}_3 $

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76.

$\vec{v} = (5,-5,2,1) $

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Answer.

$(5,-5,2,1) = 5 \vec{e}_1 + (-5) \vec{e}_2 + 2 \vec{e}_3 + 1 \vec{e}_4 $

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77.

$\vec{v} = (-6,-1,0,3) $

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Answer.

$(-6,-1,0,3) = (-6) \vec{e}_1 + (-1) \vec{e}_2 + 0 \vec{e}_3 + 3 \vec{e}_4 $

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Creating a vector parallel to a line in $\R^2 $.

In each case, create a vector that is parallel to the given line. Then draw both the graph of the line and your vector with its initial point at the origin.

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78.

$\ell \colon \;\; x - 3 y = 4 $

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Answer.

One such vector is $\vec{v} = (3,1) \text{.}$

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Diagram of a line and a vector parallel to it.

A diagram consisting of the graph of the line $x - 3 y = 4 $ in the $xy$-plane, along with the vector $\vec{v} = (3,1) $ drawn with its initial point at the origin, so that its terminal point is at $(3,1) \text{.}$

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Note: Any nonzero scalar multiple of the provided answer is another suitable answer.

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79.

$\ell \colon \;\; 2 x + 5 y = 5$

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Answer.

One such vector is $\vec{v} = (5,-2) \text{.}$

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A diagram consisting of the graph of the line $2 x + 5 y = 5 $ in the $xy$-plane, along with the vector $\vec{v} = (5,-2) $ drawn with its initial point at the origin, so that its terminal point is at $(5,-2) \text{.}$

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Note: Any nonzero scalar multiple of the provided answer is another suitable answer.

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Creating vectors parallel to a plane in $\R^3 $.

In each case, create two vectors each of which is parallel to the given plane but that are not parallel to each other.

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80.

$\ell \colon \;\; x - 5 y - 6 z = 1$

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Answer.

One such pair of vectors is $\vec{u} = (6,0,1) \text{,}$ $\vec{v} = (1,-1,1) \text{.}$

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Note: Any nonzero vector that is a linear combination of the two provided answer vectors is a suitable replacement for either of the two provided vectors, as long as the final pair of vectors are not scalar multiples of each other.

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81.

$\ell \colon \;\; 5 x + 3 y - 4 z = 2 $

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Answer.

One such pair of vectors is $\vec{u} = (-1,3,1) \text{,}$ $\vec{v} = (0,4,3) \text{.}$

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Geometry using vector calculations.

For each pair of points $P,Q \text{:}$

Use a vector calculation to determine the point that bisects segment $\absegment{PQ} \text{.}$

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Use a vector calculation to determine the two points that trisect segment $\absegment{PQ} \text{.}$

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Hint: Think of vector addition as a sequence of consecutive displacements, beginning at the origin.

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82.

$P(1,-8) \text{,}$ $Q(-7,-3) \text{.}$

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Solution.

Let $\vec{u} $ represent the vector associated to directed line segment $\abray{OP} $ and let $\vec{v} $ represent the vector associated to directed line segment $\abray{PQ} \text{,}$ so that $\vec{u} = (1,-8) $ and $\vec{v} = (-8,5) \text{.}$

The bisecting point will be the terminal point of the vector
\begin{equation*} \vec{b} = \vec{u} + \tfrac{1}{2} \vec{v} = ( -3, -11/2 ) \text{,} \end{equation*}
when placed with its initial point at the origin. That is, the bisecting point is $R(-3,-11/2) \text{.}$

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The trisecting points will be the terminal points of the vectors
\begin{align*} \vec{t}_1 \amp = \vec{u} + \tfrac{1}{3} \vec{v} = ( - 5/3, -19/3 ) \text{,} \\ \vec{t}_2 \amp = \vec{u} + \tfrac{2}{3} \vec{v} = ( -13/3, -14/3 ) \text{,} \end{align*}
when placed with their initial points at the origin. That is, the trisecting points are $R(-5/3,-19/3) $ and $S(-13/3,-14/3) \text{.}$

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83.

$P(4,0,6) \text{,}$ $Q(-2,9,-1) \text{.}$

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Solution.

The bisecting point will be the terminal point of the vector
\begin{equation*} \vec{b} = \vec{u} + \tfrac{1}{2} \vec{v} = (1,9/2,5/2) \text{,} \end{equation*}
when placed with its initial point at the origin. That is, the bisecting point is $R(1,9/2,5/2) \text{.}$

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The trisecting points will be the terminal points of the vectors
\begin{align*} \vec{t}_1 \amp = \vec{u} + \tfrac{1}{3} \vec{v} = (2,3,11/3) \text{,} \\ \vec{t}_2 \amp = \vec{u} + \tfrac{2}{3} \vec{v} = (0,6, 4/3) \text{,} \end{align*}
when placed with their initial points at the origin. That is, the trisecting points are $R(2,3,11/3) $ and $S(0,6, 4/3) \text{.}$

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84. Creating a linear combination.

Determine values for scalars $k_1, k_2, k_3 $ so that the vector equality is true.

\begin{equation*} k_1 (4,-1,-1) + k_2 (-3,1,2) + k_3 (-2,-1,-6) = (1,2,10) \end{equation*}

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Solution.

First compute

\begin{equation*} k_1 (4,-1,-1) + k_2 (-3,1,2) + k_3 (-2,-1,-6) = ( 4 k_1 - 3 k_2 - 2 k_3, - k_1 + k_2 - k_3, - k_1 + 2 k_2 - 6 k_3 )\text{.} \end{equation*}

Thus our vector equality becomes

\begin{equation*} ( 4 k_1 - 3 k_2 - 2 k_3, - k_1 + k_2 - k_3, - k_1 + 2 k_2 - 6 k_3 ) = (1,2,10) \text{.} \end{equation*}

Matching corresponding components on each side of this equality leads to a linear system of equations.

\begin{equation*} \begin{sysofeqns}{rcrcrcr} 4 k_1 \amp - \amp 3 k_2 \amp - \amp 2 k_3 \amp = \amp 1 \\ - k_1 \amp + \amp k_2 \amp - \amp k_3 \amp = \amp 2 \\ - k_1 \amp + \amp 2 k_2 \amp - \amp 6 k_3 \amp = \amp 10 \end{sysofeqns} \end{equation*}

Convert the system into an augmented matrix and reduce.

\begin{equation*} \begin{abmatrix}{rrr|r} 4 \amp -3 \amp -2 \amp 1 \\ -1 \amp 1 \amp -1 \amp 2 \\ -1 \amp 2 \amp -6 \amp 10 \end{abmatrix} \qquad\rowredarrow\qquad \begin{abmatrix}{rrr|r} 1 \amp 0 \amp 0 \amp 2 \\ 0 \amp 1 \amp 0 \amp 3 \\ 0 \amp 0 \amp 1 \amp -1 \end{abmatrix} \end{equation*}

Thus the original vector equality is true for scalar values $k_1 = 2 \text{,}$ $k_2 = 3 \text{,}$ $k_3 = -1 \text{.}$

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Determining if a vector is a linear combination of others.

In each case, determine whether the vector $\vec{w}$ can be expressed as a linear combination of the vectors $\vec{v}_1, \vec{v}_2, \dotsc, \vec{v}_m \text{.}$

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85.

$\vec{w} = ( 4, 7, 9) \text{;}$ $\vec{v}_1 = ( 0, -2, -1) \text{,}$ $\vec{v}_2 = (-1, 3, 0) \text{,}$ $\vec{v}_3 = ( 2, 7, 6) $

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Answer.

Yes,

\begin{equation*} \vec{w} = -3 \vec{v}_1 - 2 \vec{v}_2 + \vec{v}_3 \text{.} \end{equation*}

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86.

$\vec{w} = ( 2, 3, 2) \text{;}$ $\vec{v}_1 = ( 1, 1, 1) \text{,}$ $\vec{v}_2 = ( 0, 1, -1) \text{,}$ $\vec{v}_3 = ( 3, 4, 2) $

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Answer.

No.

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87.

$\vec{w} = (-1, 1, -3) \text{;}$ $\vec{v}_1 = ( 1, 1, 1) \text{,}$ $\vec{v}_2 = ( 0, 1, -1) \text{,}$ $\vec{v}_3 = ( 3, 4, 2) $

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Answer.

Yes, there are an infinite number of ways of expressing $\vec{w} $ as a linear combination of $\vec{v}_1, \vec{v}_2, \vec{v}_3 \text{.}$ The “simplest” is

\begin{equation*} \vec{w} = -7 \vec{v}_1 + 0 \vec{v}_2 + 2 \vec{v}_3 = 2 \vec{v}_3 - 7 \vec{v}_1 \text{.} \end{equation*}

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88.

$\vec{w} = ( 6, 9, 3, 11) \text{;}$ $\vec{v}_1 = ( 1, 2, 1, 2) \text{,}$ $\vec{v}_2 = ( 4, -1, -2, 3) \text{,}$ $\vec{v}_3 = ( 5, -4, -4, 2) $

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Answer.

Yes,

\begin{equation*} \vec{w} = \vec{v}_1 + 5 \vec{v}_2 - 3 \vec{v}_3 \text{.} \end{equation*}

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