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Precalculus : Matrices and Vectors

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Example Questions

Example Question #2 : Find A Vector Equation When Given Two Points

Find the vector that starts at point (3,4) and ends at (42,6) and its magnitude.

Possible Answers:

$\vec{r}= \left \langle 39,2\right \rangle$

$\left | \vec{r}\right | \approx 39.05$

$\vec{r}= \left \langle 39,-2\right \rangle$

$\left | \vec{r}\right |= 7$

$\vec{r}= \left \langle 10,45\right \rangle$

$\left | \vec{r}\right | \approx 45.11$

$\vec{r}= \left \langle 45,10\right \rangle$

$\left | \vec{r}\right | \approx 19.50$

Correct answer:

$\vec{r}= \left \langle 39,2\right \rangle$

$\left | \vec{r}\right | \approx 39.05$

Explanation:

To find the vector between two points, find the change between the points in the and directions, or $\Delta x$ and $\Delta y$ . Then $\vec{r}= \left \langle \Delta x , \Delta y\right \rangle$ . If it helps, draw a line from the starting point to the end point on a graph and look at the changes in each direction.

We see that $\Delta x = 39$ and $\Delta y = 2$ , so our vector is

$\vec{r}= \left \langle 39,2\right \rangle$

To find a vectors magnitude, we sum up the squares of each component and take the square root:

$\left | \vec{r} \right | = \sqrt{39^2+2^2} = \sqrt{1525} \approx 39.05$

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Example Question #3 : Find A Vector Equation When Given Two Points

Find the vector equation of the line through the points:

(4,-7) and (-15,27) .

Possible Answers:

<x_1,x_2>=t<19,-34>

<x_1,x_2>=<-15,27>+t<-11,-34>

<x_1,x_2>=<4,-7>+<19,-34>

<x_1,x_2>=<4,-7>+t<-15,27>

<x_1,x_2>=<4,-7>+t<19,-34>

Correct answer:

<x_1,x_2>=<4,-7>+t<19,-34>

Explanation:

The vector equation of the line through two points is the sum of one of the points and the direction vector between the two points scaled by a variable.

First we find the the direction vector by subtracting the two points:

(4,-7)-(-15,27)=<19,-34> .

Note that a line is continuous and defined on the real line. Then, we must scale the direction vector by a variable constant so as to define the line at each point. We then add one of the given points, so as to define the line through the given points. Either point can be chosen, but the correct answer uses the first point given.

<x_1,x_2>=<4,-7>+t<19,-34>

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Example Question #4 : Find A Vector Equation When Given Two Points

Given points $\textup{A}(6,9)$ and $\textup{B}(-6,-9)$ , find a vector equation of the line passing these two points.

Possible Answers:

$r=\bigl(\begin{smallmatrix} 9\\6 \end{smallmatrix}\bigr)+t\bigl(\begin{smallmatrix} -18\\-12 \end{smallmatrix}\bigr)$

$r=\bigl(\begin{smallmatrix} -6\\-9 \end{smallmatrix}\bigr)+t\bigl(\begin{smallmatrix} 12\\-18 \end{smallmatrix}\bigr)$

$r=\bigl(\begin{smallmatrix} 3\\1 \end{smallmatrix}\bigr)+t\bigl(\begin{smallmatrix} 12\\18 \end{smallmatrix}\bigr)$

$r=\bigl(\begin{smallmatrix} 0\\0 \end{smallmatrix}\bigr)+t\bigl(\begin{smallmatrix} 6\\-9 \end{smallmatrix}\bigr)$

$r=\bigl(\begin{smallmatrix} 6\\9 \end{smallmatrix}\bigr)+t\bigl(\begin{smallmatrix} -12\\-18 \end{smallmatrix}\bigr)$

Correct answer:

$r=\bigl(\begin{smallmatrix} 6\\9 \end{smallmatrix}\bigr)+t\bigl(\begin{smallmatrix} -12\\-18 \end{smallmatrix}\bigr)$

Explanation:

Write the formula to find the vector equation of the line.

r=A+tV

Using $\textup{A}(6,9)$ and $\textup{B}(-6,-9)$ , find the directional vector by subtracting point A from B.

$\left \langle -6-6,-9-9\right \rangle=\left \langle -12,-18\right \rangle$

Substitute the directional vector and point into the formula.

A possible solution is: $r=\bigl(\begin{smallmatrix} 6\\9 \end{smallmatrix}\bigr)+t\bigl(\begin{smallmatrix} -12\\-18 \end{smallmatrix}\bigr)$

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Example Question #1 : Find A Direction Vector When Given Two Points

Find the directional vector of $\underset{BA}{\rightarrow}$ if points A and B are (2,3) and (-3,2) , respectively.

Possible Answers:

$\left \langle -1,1\right \rangle$

$\left \langle \sqrt{13},\sqrt{13}\right \rangle$

$\left \langle 5,5\right \rangle$

$\left \langle -5,-1\right \rangle$

$\left \langle 5,1\right \rangle$

Correct answer:

$\left \langle 5,1\right \rangle$

Explanation:

To find vector $\underset{BA}{\rightarrow}$ , the point A is the terminal point and point B is the starting point.

The directional vector can be determined by subtracting the start from the terminal point.

$\left \langle \underset{BA}\rightarrow \right \rangle=(2,3)-(-3,2)=\left \langle 5,1\right \rangle$

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Example Question #1 : Find A Direction Vector When Given Two Points

Find the vector through the points

(12,5) and (-4,22) .

Possible Answers:

<8,27>

<16,-17>

<16,17>

<8,17>

<-16,27>

Correct answer:

<16,-17>

Explanation:

The correct vector is given by the subtraction of the two points: (12,5)-(-4,22) .

Since the subtraction here is component-wise, it is given by the formula: <x_1-x_2,y_1-y_2> .

This results in the vector <16,-17> .

The vector <-16,17> is also correct as it is a scalar multiple of the vector marked as correct, it is found by reversing the order of the subtraction of the two points.

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Example Question #1 : Find A Direction Vector When Given Two Points

Find the vector that has the initial point A(2, -2) and the terminal point B(0, 4) .

Possible Answers:

<-6, -2>

<2, -6>

<6, -2>

<-2, 6>

Correct answer:

<-2, 6>

Explanation:

To find the directional vector, subtract the coordinates of the initial point from the coordinates of the terminal point.

$\overrightarrow{AB}=<0-2, 4-(-2)>=<-2, 6>$

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Example Question #1 : Find A Direction Vector When Given Two Points

Find the direction vector with an initial point of D(-4, -10) and a terminal point C(2, 4) .

Possible Answers:

<6, 14>

<6, -14>

<14, 6>

<-6, -14>

Correct answer:

<6, 14>

Explanation:

To find the directional vector, subtract the coordinates of the initial point from the coordinates of the terminal point.

$\overrightarrow{DC}=<2-(-4), 4-(-10)>=<6, 14>$

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Example Question #1 : Find A Direction Vector When Given Two Points

Find $\overrightarrow{PQ}$ if P=(4, 1) and Q=(-5, 2) .

Possible Answers:

<-9, 1>

<6, 6>

<1, -9>

<9, -1>

Correct answer:

<-9, 1>

Explanation:

To find the direction vector going from to , subtract the x and y-coordinates of from .

$\overrightarrow{PQ}=<-5-4, 2-1>= <-9, 1>$

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Example Question #1 : Find A Direction Vector When Given Two Points

Find $\overrightarrow{MN}$ if N=(2, 2) and M=(-2, 1) .

Possible Answers:

<-4, -1>

<1, -4>

<4,1>

<4, -1>

Correct answer:

<4,1>

Explanation:

To find the direction vector from to , subtract the x- and y-coordinates of from .

$\overrightarrow{MN}=<2-(-2), 2-1>=<4, 1>$

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Example Question #1 : Find A Direction Vector When Given Two Points

Find the direction vector that has an initial point at C(1,5) and a terminal point of D(0, 1) .

Possible Answers:

<-1, -4>

<-4, 1>

<1, 4>

<-4, -1>

Correct answer:

<-1, -4>

Explanation:

To find the directional vector, subtract the coordinates of the initial point from the coordinates of the terminal point.

$\overrightarrow{CD}=<0-1, 1-5>=<-1, -4>$

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Precalculus : Matrices and Vectors

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #2 : Find A Vector Equation When Given Two Points

Example Question #3 : Find A Vector Equation When Given Two Points

Example Question #4 : Find A Vector Equation When Given Two Points

Example Question #1 : Find A Direction Vector When Given Two Points

Example Question #1 : Find A Direction Vector When Given Two Points

Example Question #1 : Find A Direction Vector When Given Two Points

Example Question #1 : Find A Direction Vector When Given Two Points

Example Question #1 : Find A Direction Vector When Given Two Points

Example Question #1 : Find A Direction Vector When Given Two Points

Example Question #1 : Find A Direction Vector When Given Two Points

All Precalculus Resources

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