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Calculus 3 : Vectors and Vector Operations

Study concepts, example questions & explanations for Calculus 3

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

Example Questions

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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Example Question #45 : Angle Between Vectors

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

u=<-3,-9>

v=<2,7>

Possible Answers:

105^o

133^o

82^o

178^o

Correct answer:

178^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

$cos(\theta)=\frac{<-3,-9>\bullet <2,7>}{\sqrt{(-3)^2+(-9)^2}*\sqrt{2^2+7^2}}=\frac{-6+(-63)}{\sqrt{90}*\sqrt{53}}$

Simplifying we get

$cos(\theta)=\frac{-69}{\sqrt{4770}}$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{-69}{\sqrt{4770}})$

and find that

$\theta=178^o$

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Example Question #46 : Angle Between Vectors

Find the angle $\theta$ in degrees between the two vectors

u=<5,-4>

v=<9,12>

Possible Answers:

57^o

92^o

81^o

35^o

Correct answer:

92^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

$cos(\theta)=\frac{<5,-4>\bullet <9,12>}{\sqrt{5^2+(-4)^2}*\sqrt{9^2+12^2}}=\frac{45+(-48)}{\sqrt{41}*\sqrt{225}}$

Simplifying we get

$cos(\theta)=\frac{-3}{\sqrt{9225}}$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{-3}{\sqrt{9225}})$

and find that

$\theta=92^o$

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Example Question #46 : Vectors And Vector Operations

Find the angle between the vectors and , where $a=\left \langle 0,0,5\right \rangle$ and $b=\left \langle 3,0,4\right \rangle$

Note: Use the dot product formula when finding the answer

Possible Answers:

$\theta=\cos^{-1}(\frac{10}{25})$

$\theta=\cos^{-1}(\frac{2}{25})$

$\theta=\cos^{-1}(\frac{20}{24})$

$\theta=\cos^{-1}(\frac{20}{25})$

Correct answer:

$\theta=\cos^{-1}(\frac{20}{25})$

Explanation:

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

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

$a\cdot b=0+0+(5*4)=20$

$\left | a\right |=\sqrt{0^2+0^2+5^2}=5$

$\left | b\right |=\sqrt{3^2+0^2+4^2}=5$

$20=25\cos(\theta)$ , and solving for theta, we get

$\theta=\cos^{-1}(\frac{20}{25})$

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Example Question #45 : Angle Between Vectors

Determine the cosine of the angle between the following vectors:

$\left \langle 2, -1, -3\right \rangle, \left \langle 0, 1, 5\right \rangle$

Possible Answers:

Correct answer:

Explanation:

The cosine of the angle, denoted by $\theta$ , between two vectors is given by the dot product of the vectors, which is the sum of the products of the corresponding components.

For our two vectors, the dot product is given by

$2(0)+1(-1)-3(5)=-16=\cos(\theta)$

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Example Question #41 : Vectors And Vector Operations

Find the angle between $\vec{A}=2i+3j+k$ and $\vec{B}=3i+4j+12k$ in degrees

Possible Answers:

Correct answer:

Explanation:

Step 1: Calculate

$A\cdot B=\begin{pmatrix} 2\\3 \\1 \end{pmatrix}\begin{pmatrix} 3\\4 \\12 \end{pmatrix}=6+12+12=30$

Step 2: Find the respective magnitudes of A and B

$A=\sqrt{2^2+3^2+1^2}=\sqrt{14}\\B=\sqrt{3^2+4^2+12^2}=13$

Step 3:

Use the formula to find the angle between two vectors and .

Let the angle between the vectors be $\theta$ . Then

$\cos\theta=\frac{a.b}{||a||||b||}$

$\cos \theta=\frac{30}{13\sqrt{14}}$

$\theta=\cos^{-1}\left[\frac{30}{13\sqrt{14}}\right]=51.92\approx52^\circ$

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Calculus 3 : Vectors and Vector Operations

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

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

Example Question #45 : Angle Between Vectors

Example Question #46 : Angle Between Vectors

Example Question #46 : Vectors And Vector Operations

Example Question #45 : Angle Between Vectors

Example Question #41 : Vectors And Vector Operations

All Calculus 3 Resources

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