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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Vectors And Vector Operations

What is the angle between the vectors $\vec{a}= \hat{i}+3\hat{j} +5\hat{k}$ and $\vec{b}= 2\hat{i}+6\hat{j}+10\hat{k}$ ?

Possible Answers:

$\theta\approx0^{\circ}$

$\theta\approx120^{\circ}$

$\theta\approx90^{\circ}$

$\theta\approx8.37^{\circ}$

Correct answer:

$\theta\approx0^{\circ}$

Explanation:

To find the angle between vectors, we must use the dot product formula

$a\cdot b=\left | a\right |\left | b\right |cos\theta$

where $a\cdot b$ is the dot product of the vectors and , respectively.

$\left | a\right |$ and $\left | b\right |$ are the magnitudes of vectors and , respectively.

$cos\theta$ is the angle between the two vectors.

Let vector be represented as $\vec{a}= a_{1}\hat{i}+a_{2}\hat{j} +a_{3}\hat{k}$ and vector be represented as $\vec{b}= b_{1}\hat{i}+b_{2}\hat{j} +b_{3}\hat{k}$ .

The dot product of the vectors and is $a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})$ .

The magnitude of vector is $\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$ and vector is $\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}$ .

Rearranging the dot product formula to solve for $\theta$ gives us $\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )$

For this problem,

$a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})=1(2)+(3)(6)+(5)(10)$

$a\cdot b=2+18+50=70$

$\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}=\sqrt{(1)^{2}+(3)^{2}+(5)^{2}}$

$\left | a\right |=\sqrt{1+9+25}=\sqrt{35}$

$\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}=\sqrt{(2)^{2}+(6)^{2}+(10)^{2}}$

$\left | b\right |=\sqrt{4+36+100}=\sqrt{140}$ $\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )=cos^{-1}\left ( \frac{70}{\sqrt{35}\sqrt{140}}\right )=cos^{-1}\left (\frac{70}{\sqrt{4900}}\right )$

$\theta=cos^{-1}\left (\frac{70}{70}\right )=cos^{-1}\left (1\right )$

$\theta\approx0^{\circ}$

The two vectors are parallel.

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Example Question #3 : Vectors And Vector Operations

Find the approximate acute angle in degrees between the vectors <2,2,5>, <0,5,-1> .

Possible Answers:

80.17

None of the other answers

79.23

12.55

29.77

Correct answer:

80.17

Explanation:

To find the angle between two vectors, use the formula

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

$\theta = \cos^{-1}(\frac{<2,2,5>\cdot<0,5,-1>}{\sqrt{4+4+25}\sqrt{0+25+1}})$

$\theta = \cos^{-1}(\frac{5}{\sqrt{858}})$

$\theta \approx 80.22$

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Example Question #8 : Angle Between Vectors

Find the angle between the following two vectors.

$\vec{A}=3\hat{i}+5\hat{j}-\hat{k}, \vec{B}=6\hat{i}-2\hat{j}+5\hat{k}$

Possible Answers:

$18.41 ^{\circ}$

$86.39 ^{\circ}$

$3.61 ^{\circ}$

$32.57 ^{\circ}$

Correct answer:

$86.39 ^{\circ}$

Explanation:

In order to find the angle between two vectors, we need to take the quotient of their dot product and their magnitudes:

$\frac{\vec{A}\cdot \vec{B}}{\left|\vec{A} \right| \left| \vec{B} \right| } = \cos{\theta}\Rightarrow \theta = \cos^{-1} \left( \frac{\vec{A}\cdot \vec{B}}{\left|\vec{A} \right| \left| \vec{B} \right| }\right )$

Therefore, we find that

$\vec{A}\cdot\vec{B}=3(6)+5(-2)-1(5)=3,$

$\left| \vec{A}\right| = \sqrt{3^2+5^2+1^2}=\sqrt{35}, \left| \vec{B}\right| = \sqrt{6^2+2^2+5^2}=\sqrt{65}$

$\theta = \cos^{-1} \left( \frac{3}{\sqrt{35}\sqrt{65} }\right)=86.39 ^{\circ}$ .

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Example Question #1 : Angle Between Vectors

Find the (acute) angle between the vectors $<1,0,\frac{1}{2}>, <\frac{3}{2},0, 12>$ in degrees.

Possible Answers:

42.690

82.341

40.542

8.659

56.310

Correct answer:

56.310

Explanation:

To find the angle between vectors, we use the formula

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

Substituting in our values, we get

$\theta = \cos^{-1}(\frac{<1,0,\frac{1}{2}> \cdot <\frac{3}{2},0,12>}{\sqrt{1^2+0^2+\frac{1}{2}^2}\sqrt{\frac{3}{2}^2+0^2+12^2}}) \approx \cos^{-1}(\frac{7.5}{13.521}) = 56.310$

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Example Question #10 : Angle Between Vectors

Find the angle between the two vectors.

u=<2,0>
$v=<\sqrt{3}, 1>$

Possible Answers:

No angle exists

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

$\theta = \frac{\pi}{3}$

$\theta = \frac{\pi}{6}$

Correct answer:

$\theta = \frac{\pi}{6}$

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=<2,0>
$v=<\sqrt{3}, 1>$ we find

$u\bullet v= 2*\sqrt3+0*1=2\sqrt3$

$\left \| u\right \|=\sqrt{2^2+0^2}=\sqrt4=2$

$\left \| v\right \|=\sqrt{\sqrt{3}^2+1^2}=\sqrt4=2$

Plugging these values in we get

$cos\, \theta = \frac{2\sqrt3}{2*2}=\frac{\sqrt3}{2}$

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

$cos^{-1}[cos(\theta)]=cos^{-1}(\frac{\sqrt3}{2})$

and find that

$\theta = \frac{\pi}{6}$

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Example Question #21 : Calculus 3

Find the angle between the two vectors.

u=<1,1>

v=<-3,3>

Possible Answers:

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

No angle exists

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

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

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 #21 : Calculus 3

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

Possible Answers:

23.448

76.552

None of the other answers

54.712

35.288

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 #12 : Angle Between Vectors

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

Possible Answers:

48.776

50.523

39.477

None of the other answers

51.224

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 #12 : Angle Between Vectors

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

Possible Answers:

28.913

78.127

71.087

None of the other answers

11.873

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 #11 : 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

1.18

1.46

1.62

2.89

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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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1 : Vectors And Vector Operations

Example Question #3 : Vectors And Vector Operations

Example Question #8 : Angle Between Vectors

Example Question #1 : Angle Between Vectors

Example Question #10 : Angle Between Vectors

Example Question #21 : Calculus 3

Example Question #21 : Calculus 3

Example Question #12 : Angle Between Vectors

Example Question #12 : Angle Between Vectors

Example Question #11 : Angle Between Vectors

All Calculus 3 Resources

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