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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 #32 : Vectors And Vector Operations

Find the angle between the vectors and if $a=\left \langle 0,1,2\right \rangle$ and $b=\left \langle 1,0,1\right \rangle$ . Hint: Do the dot product between the vectors to start.

Possible Answers:

$\cos^{-1}(\frac{2}{\sqrt{10}})$

$\cos^{-1}(\frac{2}{10})$

$\sin^{-1}(\frac{1}{5})$

Correct answer:

$\cos^{-1}(\frac{2}{\sqrt{10}})$

Explanation:

First, you must do the dot product of the vectors, because the answer choices are in terms of inverse cosine. Doing the dot product gets $a\cdot b=0+0+2=2$ . Next, you must find the magnitude of both vectors. $\left | a\right |=\sqrt{0^2+1^2+2^2}=\sqrt{5}$ and $\left | b\right |=\sqrt{1^2+0^2+1^2}=\sqrt2$ . Combining everything we have found and using the formula for the dot product, we get $2=\left | \sqrt2\right |\left | \sqrt5\right |\cos(\theta)$ . Solving for $\theta$ , we then get $\theta=\cos^{-1}(\frac{2}{{\sqrt{10}}})$ .

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Example Question #33 : Vectors And Vector Operations

Find the angle between the vectors $a=\left \langle 3,0,2\right \rangle$ and $b=\left \langle 1,2,4\right \rangle$ , given that $a\cdot b=11$ .

Possible Answers:

$\theta=\cos^{-1}(\frac{6}{\sqrt{73}})$

$\theta=\cos^{-1}(\frac{10}{4})$

$\theta=\cos^{-1}(\frac{11}{\sqrt{273}})$

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

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

Correct answer:

$\theta=\cos^{-1}(\frac{11}{\sqrt{273}})$

Explanation:

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

$a\cdot b=\left | a\right |\left | b\right |\cos\theta$ . Using this definition, we find that $\left | a\right |=\sqrt{3^2+0^2+2^2}=\sqrt{13}$ , $\left | b\right |=\sqrt{1^2+2^2+4^2}=\sqrt{21}$ . Putting what we know into the formula, we get $11=\sqrt{13}*\sqrt{21}\cos\theta$ . Solving for theta, we get $\theta=\cos^{-1}(\frac{11}{\sqrt{273}})$

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Example Question #31 : Vectors And Vector Operations

Find the angle between the vectors $a=\left \langle 4,0,2\right \rangle$ and $b=\left \langle 1,1,2\right \rangle$ , given that $\left \| a\times b\right \|=\sqrt{56}$ .

Possible Answers:

$\theta=\sin^{-1}(\frac{\sqrt{20}}{\sqrt{120}})$

$\theta=\sin^{-1}(\frac{\sqrt{56}}{\sqrt{120}})$

$\theta=\cos^{-1}(\frac{\sqrt{56}}{\sqrt{120}})$

$\theta=\sin^{-1}(\frac{\sqrt{15}}{\sqrt{120}})$

Correct answer:

$\theta=\sin^{-1}(\frac{\sqrt{56}}{\sqrt{120}})$

Explanation:

To find the angle between the vectors, we use the formula for the cross product:

$\left \| a\times b\right \|=\left | a\right |\left | b\right |\sin\theta$ . Using this definition, we find that $\left | a\right |=\sqrt{4^2+0^2+2^2}=\sqrt{20}$ , $\left | b\right |=\sqrt{1^2+1^2+2^2}=\sqrt{6}$ . Putting what we know into the formula, we get $\sqrt{56}=\sqrt{20}*\sqrt{6}\sin\theta$ . Solving for theta, we get $\theta=\sin^{-1}(\frac{\sqrt{56}}{\sqrt{120}})$

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Example Question #31 : Angle Between Vectors

Find the angle in degrees between the vectors <1,5>,<e,1> .

Possible Answers:

None of the other answers

65.447

38.992

24.553

51.008

Correct answer:

None of the other answers

Explanation:

The correct answer is about 58.492 degrees.

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

So we have

$\theta = \cos^{-1}(\frac{<1,5> \cdot <e,1>}{\sqrt{1^2+5^2}\sqrt{e^2+1^2}}) \approx \cos^{-1}(\frac{e+5}{14.7687})$

=58.492

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Example Question #40 : Vectors And Vector Operations

Find the angle in degrees between the vectors <e,e^2>, <e^2,e^4> .

Possible Answers:

61.090

77.510

28.910

None of the other answers.

12.490

Correct answer:

12.490

Explanation:

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

So we have

$\theta = \cos^{-1}(\frac{<e,e^2> \cdot <e^2,e^4>}{\sqrt{e^2+e^4}\sqrt{e^4+e^8}}) \approx \cos^{-1}(\frac{e^3+e^6}{433.7806})$

=12.490

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Example Question #41 : Angle Between Vectors

Possible Answers:

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

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

$\theta=\pi$

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

Correct answer:

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

Explanation:

Using the formula for the cross product between vectors and , we have everything but theta. Plugging in what we were given, we get:

$1=(1*2)\sin\theta$ . Solving for $\theta$ , we get $\theta=\sin^{-1}(\frac{1}{2})=\frac{\pi}{6}$

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Example Question #51 : Calculus 3

Two vectors v and w are separated by angle of 30^o. The vectors have the magnitudes:

$|\textbf{v}|=4, |\textbf{w}|=3\sqrt{3}$

What is the dot product of the two vectors $\textbf{v}\cdot\textbf{w}$ .

Possible Answers:

$6\sqrt{3}$

$12\sqrt3$

Correct answer:

Explanation:

The angle theta between two vectors v and w is defined by:

$\cos\theta=\frac{\textbf{v}\cdot\textbf{w}}{|\textbf{v}||\textbf{w}|}$

Rearranging, we can solve for the dot product:

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

Substituting the given quantities:

$\textbf{v}\cdot\textbf{w} =4\cdot 3\sqrt{3}\cdot \cos(30)=4\cdot 3\sqrt{3}\cdot\frac{\sqrt{3}}{2}=18$

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Example Question #42 : Angle Between Vectors

Find the angle between vectors $a=\left \langle 1,0,3\right \rangle$ and $b=\left \langle 2,1,1\right \rangle$ . Use the dot product when finding the solution.

Possible Answers:

$\sin^{-1}(\frac{5}{\sqrt60})$

$\cos^{-1}(\frac{5}{\sqrt{60}})$

$\cos^{-1}({\frac{4}{9}})$

$\cos^{-1}(\frac{2}{\sqrt5})$

Correct answer:

$\cos^{-1}(\frac{5}{\sqrt{60}})$

Explanation:

First, we must find the magnitude of the vectors.

$\left | a\right |=\sqrt{1^2+3^2}=\sqrt{10}$

$\left | b\right |=\sqrt{2^2+1^2+1^2}=\sqrt{6}$

Next, we find the dot product

$a\cdot b=(2*1)+(0*1)+(3*1)=5$

Plugging into the dot product formula, we get

$5=\sqrt{10*6}\cos\theta$ . Solving for theta, we then get

$\theta=\cos^{-1}(\frac{5}{\sqrt{60}})$

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Example Question #43 : Angle Between Vectors

Find the angle between vectors $a=\left \langle 2,1,3\right \rangle$ and $b=\left \langle -1,0,5\right \rangle$ . Use the dot product when finding the solution.

Possible Answers:

$\sin^{-1}({\frac{13}{\sqrt{364}}})$

$\cos^{-1}({\frac{1}{\sqrt{364}}})$

$\cos^{-1}({\frac{13}{\sqrt{6}}})$

$\cos^{-1}({\frac{13}{\sqrt{364}}})$

Correct answer:

$\cos^{-1}({\frac{13}{\sqrt{364}}})$

Explanation:

$\left | a\right |=\sqrt{2^2+1^2+3^2}=\sqrt{14}$

$\left | b\right |=\sqrt{(-1)^2+0^2+5^2}=\sqrt{26}$

Next, we find the dot product

$a\cdot b=(2*-1)+(-1*0)+(3*5)=13$

Plugging into the dot product formula, we get

$13=\sqrt{14*26}\cos\theta$ . Solving for theta, we then get

$\theta=\cos^{-1}(\frac{13}{\sqrt{364}})$

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Example Question #44 : Angle Between Vectors

Find the angle $\theta$ to the nearest degree between the two vectors

u=<4,14>

v=<-3,-2>

Possible Answers:

104^o

133^o

48^o

91^o

Correct answer:

133^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

$cos(\theta)=\frac{<4,14>\bullet <-3,-2>}{\sqrt{4^2+14^2}*\sqrt{(-3)^2+(-2)^2}}=\frac{-12+(-28)}{\sqrt{260}*\sqrt{13}}$

Simplifying we get

$cos(\theta)=\frac{-40}{\sqrt{3380}}$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{-40}{\sqrt{3380}})$

and find that

$\theta=133^o$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #32 : Vectors And Vector Operations

Example Question #33 : Vectors And Vector Operations

Example Question #31 : Vectors And Vector Operations

Example Question #31 : Angle Between Vectors

Example Question #40 : Vectors And Vector Operations

Example Question #41 : Angle Between Vectors

Example Question #51 : Calculus 3

Example Question #42 : Angle Between Vectors

Example Question #43 : Angle Between Vectors

Example Question #44 : Angle Between Vectors

All Calculus 3 Resources

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