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High School Math : Parametric, Polar, and Vector

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Example Questions

Example Question #3 : Understanding Vector Calculations

Given vector u=<0,0> and v=<2,1> , solve for 2u-3v .

Possible Answers:

<-6,-3>

<2,-3>

<-6,1>

<6,3>

Correct answer:

<-6,-3>

Explanation:

u=<0,0> v=<2,1>

To solve for 2u-3v , We need to first multiply into vector to find and multiply into vector to find ; then we need to subtract the components in the vector and the components together:

2u=<2*0,2*0> = <0,0>

3u=<3*2,3*1> = <6,3>

2u-3v= <0-6,0-3>

2u-3v= <-6,-3>

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Example Question #4 : Understanding Vector Calculations

Find the unit vector of v=<5,-12> .

Possible Answers:

$<\frac{5}{13},\frac{-12}{13}>$

$<\frac{-5}{13},\frac{-12}{13}>$

$<\frac{-5}{13},\frac{12}{13}>$

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

Correct answer:

$<\frac{5}{13},\frac{-12}{13}>$

Explanation:

To solve for the unit vector, the following formula must be used:

$\frac{v}{||v||}$

v=<5,-12>

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

$\left \| v \right \|=\sqrt{25+144}$

$\left \| v \right \|=\sqrt{169}$

$\left \| v \right \|=13$

unit vector:

$\frac{<5,-12>}{13}$

$=\frac{1}{13}*<5,-12>$

$=<\frac{5}{13},\frac{-12}{13}>$

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Example Question #1 : Vector

Is $\langle \frac{5}{13},\frac{-12}{13}\rangle$ a unit vector?

Possible Answers:

yes, because magnitude is equal to

no, because magnitude is not equal to

not enough information given

Correct answer:

yes, because magnitude is equal to

Explanation:

To verify where a vector is a unit vector, we must solve for its magnitude. If the magnitude is equal to 1 then the vector is a unit vector:

$<\frac{5}{13},\frac{-12}{13}>$

$\sqrt{\left(\frac{5}{13}\right)^2+\left(\frac{-12}{13}\right)^2}$

$\sqrt{\left(\frac{25}{169}\right)+\left(\frac{144}{169}\right)$

$\sqrt{\frac{169}{169}}$

$\sqrt{1}$

$<\frac{5}{13},\frac{-12}{13}>$ is a unit vector because magnitude is equal to .

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Example Question #3 : Understanding Vector Calculations

Given vector u=<3,3> . Solve for the direction (angle) of the vector:

Possible Answers:

$15^{\circ}$

$90^{\circ}$

$45^{\circ}$

$60^{\circ}$

Correct answer:

$45^{\circ}$

Explanation:

To solve for the direction of a vector, we use the following formula:

$tan\alpha$ = $\frac{b}{a}$

with the vector being $u=\left \langle a,b \right \rangle$

$u=\left \langle 3,3 \right \rangle$

a=3

b=3

$\tan\alpha=\frac{3}{3}=1$

$\alpha = \arctan (1)$

$\alpha =45^{\circ}$

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Example Question #1 : Understanding Vector Calculations

Solve for vector given direction of $45^{\circ}$ and magnitude of .

Possible Answers:

$\left \langle 4\sqrt{2},4\sqrt{2}} \right \rangle$

$\left \langle 2\sqrt{2},2\sqrt{2}} \right \rangle$

$\left \langle 4,4} \right \rangle$

$\left \langle 2,2} \right \rangle$

Correct answer:

$\left \langle 4\sqrt{2},4\sqrt{2}} \right \rangle$

Explanation:

To solve for a vector with the magnitude and direction given, we use the following formula:

v=(||v||cos) i + (||v||sin)j

$v=(8\cos 45^{\circ})i +(8\sin45^{\circ})j$

$v=4\sqrt{2}i+4\sqrt{2}j$

$v=\left \langle 4\sqrt{2},4\sqrt{2}} \right \rangle$

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Example Question #11 : Vector

Given vector u=<2,-1> and v=<1,2> , solve for u-2v .

Possible Answers:

<0,3>

<0,-5>

<0,-3>

<1,-3>

Correct answer:

<0,-5>

Explanation:

u=<2,-1> v=<1,2>

To solve for u-2v , We need to multiply into vector to find ; then we need to subtract the components in the vector and the components together:

2v= <2*1,2*2> = <2,4>

u-2v= <2-2,-1-4>

u-2v= <0,-5>

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Example Question #11 : Vector

Find the magnitude of v=6i-6j .

Possible Answers:

$\sqrt{24}$

$\sqrt{36}$

$6\sqrt{2}$

Correct answer:

$6\sqrt{2}$

Explanation:

v=6i-6j therefore the vector is <6,-6>

To solve for the magnitude:

$\sqrt{(6)^2+(-6)^2}$

$\sqrt{(36)+(36)}$

$\sqrt{72}$

$6\sqrt{2}$

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Example Question #22 : Calculus Ii — Integrals

Let $\vec{}$ $\vec{a}$ and $\vec{b}$ be the following vectors: $\vec{a}=\left \langle 1,2,1\right \rangle$ and $\vec{b}=\left \langle 2,3,-1\right \rangle$ . If $\theta$ is the acute angle between the vectors, then which of the following is equal to $\theta$ ?

Possible Answers:

$\cos^{-1}\left(\frac{\sqrt{21}}{14}\right)$

$\sin^{-1}\left(\frac{\sqrt{21}}{7}\right)$

$\tan^{-1}\left(\frac{\sqrt{21}}{14}\right)$

$\cos^{-1}\left(\frac{\sqrt{21}}{7}\right)$

Correct answer:

$\cos^{-1}\left(\frac{\sqrt{21}}{14}\right)$

Explanation:

The cosine of the acute angle between two vectors is given by the following formula:

$\cos\theta=\frac{\vec{a}\cdot \vec{b}}{\left \| \vec{a}\right \| \left \| \vec{b} \right \|}$ , where $\vec{a} \cdot \vec{b}$ represents the dot product of the two vectors, $\left \| \vec{a} \right \|$ is the magnitude of vector a, and $\left \| \vec{b} \right \|$ is the magnitude of vector b.

First, we will need to compute the dot product of the two vectors. Let's say we have two general vectors in space (three dimensions), $\vec{a}$ and $\vec{b}$ . Let the components of $\vec{a}$ be $<a_{1}, a_{2}, a_{3}>$ and the components of $\vec{b}$ be $<b_{1}, b_{2}, b_{3}>$ . Then the dot product $\vec{a} \cdot \vec{b}$ is defined as follows:

$\vec{a} \cdot \vec{b}=a_{1}(b_{1})+a_{2}(b_{2})+a_{3}(b_{3})$ .

Going back to the original problem, we can use this definition to find the dot product of $\vec{a}=<1,2,1>$ and $\vec{b}=<-2,3,-1>$ .

$\vec{a} \cdot \vec{b}=a_{1}(b_{1})+a_{2}(b_{2})+a_{3}(b_{3})$

=1(-2)+2(3)+1(-1)

=-2+6-1=3

The next two things we will need to compute are $\left \| \vec{a} \right \|$ and $\left \| \vec{b} \right \|$ .

Let the components of a general vector $\vec{a}$ be $<a_{1}, a_{2}, a_{3}>$ . Then $\left \| \vec{a} \right \|$ is defined as $\left \| \vec{a}\right \|=\sqrt{(a_{1})^2+(a_{2})^2+(a_3)^2}$ .

Thus, if $\vec{a}=<1,2,1>$ and $\vec{b}=<-2,3,-1>$ , then

$\left \| \vec{a}\right \|=\sqrt{(a_{1})^2+(a_{2})^2+(a_3)^2}$

$=\sqrt{(1)^2+(2)^2+(1)^2}=\sqrt{1+4+1}=\sqrt{6}$

and $\left \| \vec{b}\right \|=\sqrt{(b_{1})^2+(b_{2})^2+(b_3)^2}=\sqrt{(-2)^2+(3)^2+(-1)^2}$

$=\sqrt{4+9+1}=\sqrt{14}$ .

Now, we put all of this information together to find the cosine of the angle between the two vectors.

$\cos\theta=\frac{\vec{a}\cdot \vec{b}}{\left \| \vec{a}\right \| \left \| \vec{b} \right \|}=\frac{3}{\sqrt{6}\cdot\sqrt{14}}$

We just need to simplify this.

$\frac{3}{\sqrt{6}\cdot\sqrt{14}}=\frac{3}{\sqrt{6\cdot14}}=\frac{3}{\sqrt{2\cdot3\cdot2\cdot7}}$

$=\frac{3}{2\sqrt{21}}$ .

In order to get it completely simplified, we have to rationalize the denominator by multiplying the numerator and denominator by the sqare root of 21.

$\frac{3}{2\sqrt{21}}=\frac{3\cdot\sqrt{21}}{2\sqrt{21}\cdot\sqrt{21}}=\frac{3\sqrt{21}}{2(21)}$

$=\frac{\sqrt{21}}{2\cdot7}=\frac{\sqrt{21}}{14}$ .

We just have one more step. We need to solve for the value of the angle. In order to do this, we can take the inverse cosine of both sides of the equation.

$\cos\theta=\frac{\sqrt{21}}{14}$

$\theta=\cos^{-1}(\frac{\sqrt{21}}{14})$ .

The answer is $\cos^{-1}(\frac{\sqrt{21}}{14})$ .

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High School Math : Parametric, Polar, and Vector

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #3 : Understanding Vector Calculations

Example Question #4 : Understanding Vector Calculations

Example Question #1 : Vector

Example Question #3 : Understanding Vector Calculations

Example Question #1 : Understanding Vector Calculations

Example Question #11 : Vector

Example Question #11 : Vector

Example Question #22 : Calculus Ii — Integrals

All High School Math Resources

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