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Calculus 3 : Angle between Vectors

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

Example Questions

Example Question #31 : Vectors And Vector Operations

Find the direction angles of the vector $i-j+\sqrt{2}k$

Possible Answers:

$\alpha=\frac{\pi}{3}, \beta=\frac{2\pi}{3}, \gamma=\frac{\pi}{4}$

None of the Above

$\alpha=\frac{\pi}{3}, \beta=\frac{\pi}{3}, \gamma=\frac{3\pi}{4}$

$\alpha=\frac{2\pi}{3}, \beta=-\frac{\pi}{3}, \gamma=\frac{3\pi}{4}$

$\alpha=\frac{\pi}{3}, \beta=\frac{\pi}{3}, \gamma=\frac{\pi}{4}$

Correct answer:

$\alpha=\frac{\pi}{3}, \beta=\frac{2\pi}{3}, \gamma=\frac{\pi}{4}$

Explanation:

To find the direction angles we must first find the Unit vector of $i-j+\sqrt{2}k$ .

$\left \| (1,-1,\sqrt{2}) \right \|=\sqrt{1+1+2}=\sqrt{4}=2$

$\vec{u}=(\frac{1}{2},-\frac{1}{2},\frac{\sqrt{2}}{2})$

Then we use Cosine to find each angle:

$\cos \alpha=\frac{1}{2}, \cos \beta=-\frac{1}{2}, \cos \gamma=\frac{\sqrt{2}}{2}$

so,

$\alpha=\frac{\pi}{3}, \beta=\frac{2\pi}{3}, \gamma=\frac{\pi}{4}$

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Example Question #32 : Vectors And Vector Operations

If a = (3,2,−1) and b = (6,α,−2) are parallel, then α =

Possible Answers:

None of the Above

Correct answer:

Explanation:

If a and b are parallel, then there is a scaler multiple of $\vec{b}=\beta \vec{a}$ :

in this case $\beta=2$ . Therefore,

$\vec{b}=2\vec{a}$

so,

$\alpha=2*2=4$

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Example Question #33 : Vectors And Vector Operations

Find the angle between the vectors

u=<3,17>

v=<5,2>

Round to the nearest tenth.

Possible Answers:

$\theta=34.1^o$

$\theta=81.3^o$

$\theta=104.9^o$

$\theta=58.2^o$

Correct answer:

$\theta=58.2^o$

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

$cos(\theta)=\frac{3(5)+17(2)}{\sqrt{3^2+17^2}*\sqrt{5^2+2^2}}$

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

$\theta=58.2^o$

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Example Question #31 : Vectors And Vector Operations

Find the angle between the vectors and , given that $\left | a\right |=1$ , $\left | b\right |=2$ , and $a\cdot b=\sqrt2$ .

Possible Answers:

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

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

$\theta=2\pi$

$\theta=\pi$

Correct answer:

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

Explanation:

Using the dot product formula $a\cdot b=\left | a\right |\left | b\right |\cos(\theta)$ . Plugging in what we were given in the problem statement, we get $\sqrt2=2\cos(\theta)$ . Solving for $\theta$ we get $\cos^{-1}(\frac{\sqrt2}{2})=\theta=\frac{\pi}{4}$ .

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Example Question #35 : Vectors And Vector Operations

Find the angle between the vectors and , given that $\left | a\right |=1$ , $\left | b\right |=1$ , and $a\cdot b= -1$ .

Possible Answers:

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

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

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

$\theta=\pi$

Correct answer:

$\theta=\pi$

Explanation:

Using the dot product formula $a\cdot b=\left | a\right |\left | b\right |\cos(\theta)$ . Plugging in what we were given in the problem statement, we get $-1=1\cos(\theta)$ . Solving for $\theta$ we get $\cos^{-1}(-1)=\theta=\pi$ .

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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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Calculus 3 : Angle between Vectors

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #31 : Vectors And Vector Operations

Example Question #32 : Vectors And Vector Operations

Example Question #33 : Vectors And Vector Operations

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

Example Question #31 : Vectors And Vector Operations

Example Question #35 : Vectors And Vector Operations

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

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