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Calculus 3 : Angle between Vectors

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Example Questions

Example Question #21 : Vectors And Vector Operations

Let $\vec{A}$ be any arbitrary real valued vector inclined at an angle $30\degree$ to the horizontal. Calculate the projection of the vector on the horizontal.

Possible Answers:

$\vec{A}\cos 30\degree$

$\vec{A}\sec 30\degree$

$\vec{A}\sin 30\degree$

$\vec{A}\tan 30\degree$

Correct answer:

$\vec{A}\cos 30\degree$

Explanation:

Projection means a shadow. If light is cast on the vector from above, it will cast a shadow on the horizontal plane.

Untitled

We know

$\cos\theta=\frac{adj.}{hyp.}$

The hypotenuse here is the vector $\vec{A}$ itself. Solving for the adjacent side gives us the horizontal projection to be $\vec{A}\cos 30\degree$

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Example Question #21 : Vectors And Vector Operations

Find the angle between the vectors $\left \langle 3,0,0\right \rangle$ and $\left \langle 3,1,0\right \rangle$ .

Possible Answers:

$\cos^{-1}(\frac{3}{\sqrt10})$

$\cos^{-1}(\frac{9}{10})$

$\frac{1}{2}$

$\frac{\sqrt2}{2}$

Correct answer:

$\cos^{-1}(\frac{3}{\sqrt10})$

Explanation:

The formula for finding the angle between the vectors is the dot product formula, which is $a\cdot b=\left | a\right |\left | b\right |\cos\theta$ . First, we find the value of $a\cdot b=(3*3)+(1*0)+(0*0)=9$ . Next we find the magnitude of both a and b. $\left | a\right |=\sqrt(3^2+0+0)=3$ and $\left | b\right |=\sqrt(3^2+1^2+0)=\sqrt10$ . Plugging in and solving for theta, we get $\cos(\theta)=\frac{3}{\sqrt10}$ . To get theta by itself, we take the inverse cosine of both sides.

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Example Question #23 : Vectors And Vector Operations

Find the angle between the vectors a and b, where $\left | a\right |=4$ , $\left | b\right |=2$ , and $a\times b=4$

Possible Answers:

$\frac{\pi}{6}$

$\frac{\pi}{}3$

$\pi$

Correct answer:

$\frac{\pi}{6}$

Explanation:

Using the formula for the cross product, which is $a\times b=\left | a\right |\left | b\right |\sin(\theta)$ , we have all the values except theta. Plugging in the known values and solving, we get $\sin(\theta)=\frac{1}{2}$ . Therefore, $\theta=\sin^{-1}(\frac{1}{2})=\frac{\pi}{6}$

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Example Question #24 : Vectors And Vector Operations

The angle between a = (2, −1,1) and b = (−1, 2,1) is:

Possible Answers:

$\frac{\pi}{3}$

None of the Above

$\frac{5\pi}{6}$

$\frac{3\pi}{4}$

$\frac{2\pi}{3}$

Correct answer:

$\frac{2\pi}{3}$

Explanation:

In order to find the angle between two vectors we use:

$\cos\theta=\frac{\vec{a}\cdot \vec{b}}{\left \| \vec{a} \right \|\left \| \vec{b} \right \|}$

$\left \| \vec{a} \right \|=\sqrt{6}$ and $\left \| \vec{b} \right \|=\sqrt{6}$

$\vec{a}\cdot \vec{b}=-2-2+1=-3$

so,

$\cos \theta=\frac{-3}{6}=\frac{-1}{2}$

Therefore,

$\theta=\frac{2\pi}{3}$

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Example Question #25 : Vectors And Vector Operations

Find the angle in degrees between the two vectors. Round the answer to the nearest tenth.

u=<4,2>

v=<8,13>

Possible Answers:

$\theta=40.5^o$

$\theta=50.7^o$

$\theta=20.3^o$

$\theta=31.8^o$

Correct answer:

$\theta=31.8^o$

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

$cos(\theta)=\frac{4*8+2*13}{\sqrt{4^2+2^2}*\sqrt{8^2+13^2}}$

Simplifying we get

$cos(\theta)=\frac{58}{\sqrt{4660}}$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{58}{\sqrt{4660}})$

and find that

$\theta=31.8^o$

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Example Question #26 : Vectors And Vector Operations

Find the angle between the two vectors.

u=<-3,6>

v=<5,7>

Possible Answers:

$\theta=81.3^o$

$\theta=96.6^o$

$\theta=42.5^o$

$\theta=62.1^o$

Correct answer:

$\theta=62.1^o$

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

$cos(\theta)=\frac{-3*5+6*7}{\sqrt{(-3)^2+6^2}*\sqrt{5^2+7^2}}$

Simplifying we get

$cos(\theta)=\frac{27}{\sqrt{3330}}$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{27}{\sqrt{3330}})$

and find that

$\theta=62.1^o$

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Example Question #27 : Vectors And Vector Operations

Find the angle between vectors and .

u=<10,20>

v=<7,10>

Round your answer to the nearest degree.

Possible Answers:

$\theta=102.5^o$

$\theta=34.6^o$

$\theta=57.1^o$

$\theta=8.4^o$

Correct answer:

$\theta=8.4^o$

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

$cos(\theta)=\frac{10*7+20*10}{\sqrt{10^2+20^2}*\sqrt{7^2+10^2}}$

Simplifying we get

$cos(\theta)=\frac{270}{\sqrt{74500}}$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{270}{\sqrt{74500}})$

and find that

$\theta=8.4^o$

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Example Question #28 : Vectors And Vector Operations

Find the angle between vectors a and b, where $\left | a\right |=4$ , $\left | b\right |=\frac{1}{2}$ , and $a\cdot b=\sqrt2$ .

Possible Answers:

$\pi$

$\frac{\pi}{2}$

$2\pi$

$\frac{\pi}{4}$

Correct answer:

$\frac{\pi}{4}$

Explanation:

Using the formula for the dot product, which is

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

all we do not have is the value of theta.

Plugging in what we know, we get

$\cos\theta=\frac{\sqrt2}{2}$ .

Doing inverse cosine of both sides gets us

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

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Example Question #29 : Vectors And Vector Operations

Find the angle between the vectors a and b, where $\left | a\right |=4$ , $\left | b\right |=\frac{1}{2}$ , and $a\times b=\sqrt3$ .

Possible Answers:

$\pi$

$\frac{\pi}{4}$

$\frac{\pi}{2}$

$\frac{\pi}{3}$

Correct answer:

$\frac{\pi}{3}$

Explanation:

Using the formula for the cross product, which is

$a\times b=\left | a\right |\left | b\right |\sin(\theta)$ ,

we have all the values except theta.

Plugging in the known values and solving, we get

$\sin(\theta)=\frac{\sqrt3}{2}$ .

Therefore,

$\theta=\sin^{-1}(\frac{\sqrt3}{2})=\frac{\pi}{3}$

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Example Question #30 : Vectors And Vector Operations

Find the angle between the vectors a and b, where $\left | a\right |=4$ , $\left | b\right |=\frac{1}{2}$ , and $a\cdot b=\sqrt3$ .

Possible Answers:

$\frac{\pi}{6}$

$\pi$

$\frac{\pi}{3}$

Correct answer:

$\frac{\pi}{6}$

Explanation:

Using the formula for the dot product, which is

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

we have all the values except theta.

Plugging in the known values and solving, we get

$\cos(\theta)=\frac{\sqrt3}{2}$ .

Therefore,

$\theta=\cos^{-1}(\frac{\sqrt3}{2})=\frac{\pi}{6}$

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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 : Vectors And Vector Operations

Example Question #21 : Vectors And Vector Operations

Example Question #23 : Vectors And Vector Operations

Example Question #24 : Vectors And Vector Operations

Example Question #25 : Vectors And Vector Operations

Example Question #26 : Vectors And Vector Operations

Example Question #27 : Vectors And Vector Operations

Example Question #28 : Vectors And Vector Operations

Example Question #29 : Vectors And Vector Operations

Example Question #30 : Vectors And Vector Operations

All Calculus 3 Resources

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