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Calculus 3 : Angle between Vectors

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Example Questions

Example Question #21 : Calculus 3

Find the angle between the two vectors.

u=<1,1>

v=<-3,3>

Possible Answers:

No angle exists

$\theta = \frac{\pi}{8}$

$\theta = \frac{5\pi}{4}$

$\theta = \frac{\pi}{2}$

Correct answer:

$\theta = \frac{\pi}{2}$

Explanation:

To find the angle between two vector we use the following formula

$cos \, \theta = \frac{u\bullet v}{\left \| u\right \|*\left \| v\right \|}$

and solve for $\theta$ .

Given

u=<1,1>

v=<-3,3>

we find

$u\bullet v= 1*(-3)+1*3=0$

$\left \| u\right \|=\sqrt{1^2+1^2}=\sqrt2$

$\left \| v\right \|=\sqrt{(-3)^2+3^2}=\sqrt{18}=3\sqrt2$

Plugging these values in we get

$cos\, \theta = \frac{0}{3\sqrt2*\sqrt2}=0$

To find $\theta$ we calculate the $cos^{-1}$ of both sides

$cos^{-1}[cos(\theta)]=cos^{-1}(0)$

and find that

$\theta = \frac{\pi}{2}$

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Example Question #22 : Calculus 3

Find the approximate angle in degrees between the vectors $<1, e, e^2>, <e, \sqrt{e}, \sqrt[3]{e}>$ .

Possible Answers:

54.712

76.552

None of the other answers

35.288

23.448

Correct answer:

54.712

Explanation:

We can compute the (acute) angle between the two vectors using the formula

$\theta = \cos^{-1}(\frac{\mathbf{a}\cdot\mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|})$

Hence we have

$\theta = \cos^{-1}(\frac{<1,e,e^2>\cdot<e,\sqrt{e},\sqrt[3]{e}>}{\sqrt{1+e^2+e^4}\sqrt{e^2+e+e^{3/2}}})$

$\approx \cos^{-1}(\frac{e+e\sqrt{e}+e^{7/3}}{30.314})$

$\approx \cos^{-1}(\frac{17.512}{30.314})$

$\approx 54.712$

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Example Question #13 : Angle Between Vectors

Find the angle in degrees between the vectors $<1,1>, <\pi^e,e^{\pi}>$ .

Possible Answers:

48.776

None of the other answers

39.477

51.224

50.523

Correct answer:

None of the other answers

Explanation:

The correct answer is approximately 0.859 degrees.

To find the angle between two vectors, we use the equation $\theta = \cos^{-1}(\frac{\mathbf{a}\cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|})$ .

Hence we have

$\theta = \cos^{-1}(\frac{(1)(\pi^e)+(1)(e^{\pi})}{\sqrt{1^2+1^2}\sqrt{\pi^{2e}+e^{2\pi}}})$

= .859

(This answer is small due to the fact that the two vectors nearly point in the same direction, due to $e^\pi$ and $\pi^e$ being close in value.)

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Example Question #23 : Calculus 3

Find the acute angle in degrees between the vectors <10,1,5>, <1,-1,-1> .

Possible Answers:

78.127

None of the other answers

71.087

11.873

28.913

Correct answer:

78.127

Explanation:

To find the angle between two vectors, we use the formula

$\theta = \cos^{-1}(\frac{\mathbf{a}\cdot\mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|})$ .

So we have

$\theta = \cos^{-1}(\frac{<10,1,5>\cdot<1,-1,-1>}{\sqrt{10^2+1^2+5^2}\sqrt{1^2+(-1)^2+(-1)^2}})$

$\theta = \cos^{-1}(\frac{4}{\sqrt{126}\sqrt{3}})$

$\theta \approx 78.127$

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Example Question #12 : Angle Between Vectors

Find the angle between the following vectors (to two decimal places):

$\mathbf{v}=[-2, 4, -1],\;\mathbf{w}=[3, 1, -5]$

Possible Answers:

0.98

2.89

1.62

1.18

1.46

Correct answer:

1.46

Explanation:

The dot product is defined as:

$\mathbf{v}\cdot \mathbf{w}=|\mathbf{v}||\mathbf{w}|\cos\theta$

Where theta is the angle between the two vectors. Solving for theta:

$\theta=\cos^{-1}(\frac{\mathbf{v}\cdot \mathbf{w}}{|\mathbf{v}||\mathbf{w}|})$

To solve each component:

$\mathbf{v}\cdot \mathbf{w}=(3)(-2)+(1)(4)+(-5)(-1)=-6+4+5=3$

$|\mathbf{v}|=\sqrt{3^2+1^2+(-5)^2}=\sqrt{9+1+25}=\sqrt{35}$

$|\mathbf{w}|=\sqrt{(-2)^2+4^2+(-1)^2}=\sqrt{4+16+1}=\sqrt{21}$

Putting it all together, we can solve for theta:

$\theta=\cos^{-1}(\frac{3}{\sqrt{35}\sqrt{21}})\approx 1.46$

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Example Question #16 : Angle Between Vectors

Find the angle between vectors $u= \left \langle 1,-2,3\right \rangle$ and $v=\left \langle -3,-4,-6\right \rangle$ and round to the nearest degree.

Possible Answers:

148

132

128

116

Correct answer:

116

Explanation:

Write the formula to find the angle between two vectors.

$\theta= cos^{-1}(\frac{u_1v_1+u_2v_2+u_3v_3}{\left | u\right |\left | v\right |})$

Evaluate each term.

u_1v_1 = 1(-3)=-3

u_2v_2 = (-2)(-4)=8

u_3v_3 = 3(-6) = -18

u_1v_1+u_2v_2+u_3v_3= -13

$\left | u\right | = \sqrt{1^2+(-2)^2+(3)^2} = \sqrt{1+4+9} = \sqrt{14}$

$\left | v\right | = \sqrt{(-3)^2+(-4)^2+(-6)^2} = \sqrt{9+16+36}= \sqrt{61}$

$\left | u\right |\left | v\right | = \sqrt{14}\sqrt{61} = \sqrt{ 854}$

Substitute the values into the equation.

$\theta= cos^{-1}(\frac{-13}{\sqrt{854}})= 116.41379$

The correct answer is: 116

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Example Question #17 : Angle Between Vectors

Find the angle between the two vectors

u=<3,7>

v=<-7,10>

Possible Answers:

$\theta=70^o$

$\theta=58^o$

$\theta=90^o$

$\theta=25^o$

Correct answer:

$\theta=58^o$

Explanation:

In order to find the angle between the two vectors, we follow the formula

$cos(\theta)=\frac{u\bullet v}{||u||*||v||}$

and solve for $\theta$ .

Using the vectors in the problem, we get

$cos(\theta)=\frac{<3,7>\bullet <-7,10>}{\sqrt{3^2+7^2}*\sqrt{(-7)^2+10^2}}=\frac{3(-7)+7(10)}{\sqrt{58}*\sqrt{149}}$

Simplifying we get

$cos(\theta)=\frac{49}{\sqrt{8642}}$

To solve for $\theta$ we find the $cos^{-1}$ of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{49}{\sqrt{8642}})$

and find that

$\theta=58^o$

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Example Question #18 : Angle Between Vectors

What is the angle to the nearest degree between the vectors <4,5> and <3,-7> ?

Possible Answers:

$\theta=118^o$

$\theta=143^o$

$\theta=67^o$

$\theta=35^o$

Correct answer:

$\theta=118^o$

Explanation:

In order to find the angle between the two vectors, we follow the formula

$cos(\theta)=\frac{u\bullet v}{||u||*||v||}$

and solve for

$\theta$

Using the vectors in the problem, we get

$cos(\theta)=\frac{<4,5>\bullet <3,-7>}{\sqrt{4^2+5^2}*\sqrt{3^2+(-7)^2}}=\frac{4*3+5*(-7)}{\sqrt{41}\sqrt{58}}$

Simplifying we get

$cos(\theta)=\frac{-23}{\sqrt{2378}}$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{-23}{\sqrt{2378}})$

and find that

$\theta=118^o$

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Example Question #19 : Angle Between Vectors

Find the angle between the two vectors

<3,4> and <4,3>

Round to the nearest degree.

Possible Answers:

$\theta=10^o$

$\theta=16^o$

$\theta=54^o$

$\theta=30^o$

Correct answer:

$\theta=16^o$

Explanation:

In order to find the angle between the two vectors, we follow the formula

$cos(\theta)=\frac{u\bullet v}{||u||*||v||}$

and solve for

$\theta$

Using the vectors in the problem, we get

$cos(\theta)=\frac{<3,4>\bullet <4,3>}{\sqrt{3^2+4^2}*\sqrt{4^2+3^2}}=\frac{3*4+4*3}{\sqrt{25}\sqrt{25}}$

Simplifying we get

$cos(\theta)=\frac{24}{25}$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{24}{25})$

and find that

$\theta=16^o$

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Example Question #20 : Angle Between Vectors

Find the angle between the two vectors

u=<5,0>

v=<-3,-3>

Possible Answers:

$\theta=135^o$

$\theta=90^o$

$\theta=60^o$

$\theta=45^o$

Correct answer:

$\theta=135^o$

Explanation:

In order to find the angle between the two vectors, we follow the formula

$cos(\theta)=\frac{u\bullet v}{||u||*||v||}$

and solve for $\theta$ .

Using the vectors in the problem, we get

$cos(\theta)=\frac{<5,0>\bullet <-3,-3>}{\sqrt{5^2+0^2}*\sqrt{(-3)^2+(-3)^2}}=\frac{5(-3)+0(-3)}{\sqrt{25}*\sqrt{18}}$

Simplifying we get

$cos(\theta)=\frac{-15}{\sqrt{450}}$

To solve for $\theta$ we find the $cos^{-1}$ of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{-15}{\sqrt{450}})$

and find that

$\theta=135^o$

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Calculus 3 : Angle between Vectors

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #21 : Calculus 3

Example Question #22 : Calculus 3

Example Question #13 : Angle Between Vectors

Example Question #23 : Calculus 3

Example Question #12 : Angle Between Vectors

Example Question #16 : Angle Between Vectors

Example Question #17 : Angle Between Vectors

Example Question #18 : Angle Between Vectors

Example Question #19 : Angle Between Vectors

Example Question #20 : Angle Between Vectors

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