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Precalculus : Algebraic Vectors and Parametric Equations

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Example Questions

Example Question #1 : Find The Unit Vector In The Same Direction As A Given Vector

Find the unit vector that is in the same direction as the vector $\vec{v}= [3,6,1]$

Possible Answers:

$\vec{u}_v= \left [ {\frac{3}{\sqrt{46}}},\frac{6}{\sqrt{46}} , \frac{1}{\sqrt{46}} \right ]$

$\vec{u}_v= \left [ 9,36,1 \right ]$

$\vec{u}_v= \left [ 9,3,1 \right ]$

$\vec{u}_v= \left [ {\frac{9}{\sqrt{3}}},\frac{36}{\sqrt{6}} ,1 \right ]$

Correct answer:

$\vec{u}_v= \left [ {\frac{3}{\sqrt{46}}},\frac{6}{\sqrt{46}} , \frac{1}{\sqrt{46}} \right ]$

Explanation:

To find the unit vector in the same direction as a vector, we divide it by its magnitude.

The magnitude of $\vec{v}$ is $\left | \vec{v} \right |= \sqrt{3^2+6^2+1^2}=\sqrt{46}$ .

We divide vector $\vec{v}$ by its magnitude to get the unit vector $\vec{u}_v$ :

$\vec{u}_v= \frac{\vec{v}}{\left | \vec{v} \right |}=\frac{1}{\sqrt{46}}\cdot[3,6,1]$

$\vec{u}_v= \left [ {\frac{3}{\sqrt{46}}},\frac{6}{\sqrt{46}} , \frac{1}{\sqrt{46}} \right ]$

All unit vectors have a magnitude of , so to verify we are correct:

$\left | \vec{u}_v \right | = \sqrt{ (\frac{3}{\sqrt{46}})^2 + (\frac{6}{\sqrt{46}})^2 + (\frac{9}{\sqrt{46}})^2 } = \sqrt{\frac{9}{46} + \frac{36}{46} + \frac{1}{46} } = 1$

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Example Question #1 : Find The Unit Vector In The Same Direction As A Given Vector

A unit vector has length .

Given the vector

v=<3,6,44,-5,12>

find the unit vector in the same direction.

Possible Answers:

$<\frac{3}{44},\frac{3}{22},1,\frac{-5}{44},\frac{3}{11}>$

<3,6,44,-5,12>

$<1,2,\frac{44}{3},\frac{-5}{3},4>$

<0.065,0.129,0.949,-0.108,0.259>

$<\frac{1}{2},1,\frac{22}{3},\frac{-5}{6},2>$

Correct answer:

<0.065,0.129,0.949,-0.108,0.259>

Explanation:

First, you must find the length of the vector. This is given by the equation:

$\sqrt{3^2+6^2+44^2+(-5)^2+12^2}=\left \| v\right \|$

Then, dividing the vector by its length gives the unit vector in the same direction.

$\frac{v}{\left \| v\right \|}=\frac{<3,6,44,-5,12>}{\sqrt{2150}}\\ \\ \approx<0.065,0.129,0.949,-0.108,0.259>$

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Example Question #1 : Find The Unit Vector In The Same Direction As A Given Vector

Put the vector $\small (1,3)$ in unit vector form.

Possible Answers:

$\small (3,1)$

$\left (\frac{1}{10},\frac{3}{10}\right)$

$\left (\frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right)$

$\left (\frac{3}{10},\frac{1}{10}\right)$

$\left (\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right)$

Correct answer:

$\left (\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right)$

Explanation:

To get the unit vector that is in the same direction as the original vector $\small v$ , we divide the vector by the magnitude of the vector.

For $\small (1,3)$ , the magnitude is:

$\sqrt{x^2+y^2}$

$\small \sqrt{1^2+3^2}=\sqrt{10}$ .

This means the unit vector in the same direction of $\small (1,3)$ is,

$\small \frac{1}{\sqrt{10}}(1,3)=\left(\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right)$ .

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Example Question #1 : Find The Unit Vector In The Same Direction As A Given Vector

Find the unit vector of

$\left<3,\,4\right>$ .

Possible Answers:

$\left<1,3\right>$

$\left<1,1\right>$

$\left<1,0\right>$

$\left<\frac{3}{5},\, \frac{4}{5}\right>$

Correct answer:

$\left<\frac{3}{5},\, \frac{4}{5}\right>$

Explanation:

In order to find the unit vector u of a given vector v, we follow the formula

$u=\frac{v}{\left | v\right |}$

Let

$v=\left<a,b\right>$

The magnitude of v follows the formula

$\left | v\right |=\sqrt{a^{2}+b^{2}}$ .

For this vector in the problem

$\left | v\right |=\sqrt{3^{2}+4^{2}}$

$=\sqrt{9+16}$

$=\sqrt{25}$

$\left | v\right |=5$ .

Following the unit vector formula and substituting for the vector and magnitude

$u=\frac{\left< 3,4\right>}{5}$ .

As such,

$u\,=\,\left<\frac{3}{5},\, \frac{4}{5}\right>$ .

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Example Question #1 : Find The Unit Vector In The Same Direction As A Given Vector

Find the unit vector of

$\left<5,\,12\right>$

Possible Answers:

$\left<1, \frac{12}{5}\right>$

$\left<\frac{1}{2},\frac{1}{2}\right>$

$\left<2,1\right>$

$\left<\frac{5}{13},\,\frac{12}{13}\right>$

Correct answer:

$\left<\frac{5}{13},\,\frac{12}{13}\right>$

Explanation:

In order to find the unit vector u of a given vector v, we follow the formula

$u=\frac{v}{\left | v\right |}$

Let

$v=\left<a,b\right>$

The magnitude of v follows the formula

$\left | v\right |=\sqrt{a^{2}+b^{2}}$

For this vector in the problem

$\left | v\right |=\sqrt{5^{2}+12^{2}}$

$=\sqrt{25+144}$

$=\sqrt{169}$

$\left | v\right |=13$

Following the unit vector formula and substituting for the vector and magnitude

$u=\frac{\left< 5,12\right>}{13}$

As such,

$u\,=\,\left<\frac{5}{13},\, \frac{12}{13}\right>$

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

What is the vector that connects the point (2,9) to (7, 18) ?

Possible Answers:

$10.3, \measuredangle 60.9^{\circ}$

$11.4, \measuredangle 58.4^{\circ}$

$12.1, \measuredangle 64.3^{\circ}$

$8.7, \measuredangle 57.2^{\circ}$

None of the other answers

Correct answer:

$10.3, \measuredangle 60.9^{\circ}$

Explanation:

The first step when solving for a vetor is to find the length of the vector. It can be helpful to visualize the system as a right triangle as seen below:

At this point, we can essentially solve the problem as if we're finding the hypotenuse and angle of a right triangle.

Using the Pythagorean Theorem to find the length of the vector we get:

$v = \sqrt{5^2+9^2} = \sqrt{106} = 10.3$

Now we need to find the angle of the vector. We have all three sides of the triangle, so use the trig function that you are most comfortable with. The tangent function is used below because it uses sides that were given in the problem statement.

$tan(\theta) = \frac{9}{5} = 1.8$

$\theta = tan^{-1}(1.8) = 60.9^{\circ}$

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Example Question #1 : 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:

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

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

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

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

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

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

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

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

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:

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

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

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

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

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

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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Precalculus : Algebraic Vectors and Parametric Equations

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Find The Unit Vector In The Same Direction As A Given Vector

Example Question #1 : Find The Unit Vector In The Same Direction As A Given Vector

Example Question #1 : Find The Unit Vector In The Same Direction As A Given Vector

Example Question #1 : Find The Unit Vector In The Same Direction As A Given Vector

Example Question #1 : Find The Unit Vector In The Same Direction As A Given Vector

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

Example Question #1 : 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

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