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Calculus 3 : Vectors and Vector Operations

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Example Questions

Example Question #51 : Vectors And Vector Operations

Find the angle between the two vectors. Round to the nearest degree.

u=<4,8>

v=<7,-7>

Possible Answers:

$\theta=54^o$

$\theta=134^o$

$\theta=22^o$

$\theta=108^o$

Correct answer:

$\theta=108^o$

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

$cos(\theta)=\frac{<4,8>\bullet <7,-7>}{\sqrt{4^2+8^2}*\sqrt{7^2+(-7)^2}}=\frac{4(7)+8(-7)}{\sqrt{80}*\sqrt{98} }$

Simplifying we get

$cos(\theta)=\frac{-28}{\sqrt{7840}}$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{-28}{\sqrt{7840}})$

and find that

$\theta=108^o$

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Example Question #51 : Vectors And Vector Operations

Find the angle between the two vectors. Round to the nearest degree.

u=<9,-5>

v=<5,9>

Possible Answers:

$\theta=135^o$

$\theta=60^o$

$\theta=90^o$

$\theta=45^o$

Correct answer:

$\theta=90^o$

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

$cos(\theta)=\frac{<9,-5>\bullet <5,9>}{\sqrt{9^2+(-5)^2}*\sqrt{5^2+9^2}}=\frac{9(5)+(-5)9}{\sqrt{106}*\sqrt{106} }$

Simplifying we get

$cos(\theta)=0$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(0)$

and find that

$\theta=90^o$

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Example Question #61 : Calculus 3

Find the angle between the two vectors. Round to the nearest degree.

u=<1,-8>

v=<0,6>

Possible Answers:

$\theta=81^o$

$\theta=71^o$

$\theta=173^o$

$\theta=116^o$

Correct answer:

$\theta=173^o$

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

$cos(\theta)=\frac{<1,-8>\bullet <0,6>}{\sqrt{1^2+(-8)^2}*\sqrt{0^2+6^2}}=\frac{1(0)+(-8)6}{\sqrt{65}*\sqrt{36} }$

Simplifying we get

$cos(\theta)=\frac{-48}{\sqrt{2340}}$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{-48}{\sqrt{2340}})$

and find that

$\theta=173^o$

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Example Question #53 : Vectors And Vector Operations

Calculate the angle between the vectors $\mathbf a = \left \langle 1,-1,0\right \rangle$ and $\mathbf b = \left \langle 3,0,0\right \rangle$ , and express the measurement of the angle in degrees.

Possible Answers:

$30^\circ$

$60^\circ$

$135^\circ^{}$

$90^\circ$

$45^\circ$

Correct answer:

$45^\circ$

Explanation:

The angle $\Theta$ between the vectors $\mathbf a = \left \langle a_1,a_2,a_3\right \rangle$ and $\mathbf b = \left \langle b_1,b_2,b_3\right \rangle$ is given by the following equation:

$\cos\Theta=\frac{\mathbf a \cdot \mathbf b}{|\mathbf a| \ |\mathbf b|}$

where $\mathbf a \cdot \mathbf b$ represents the cross product of the vectors $\mathbf a$ and $\mathbf b$ , and $|\mathbf a|$ and $|\mathbf b|$ represent the respective magnitudes of the vectors $\mathbf a$ and $\mathbf b$ .

We are given the vectors $\mathbf a = \left \langle 1,-1,0\right \rangle$ and $\mathbf b = \left \langle 3,0,0\right \rangle$ . Calculate $\mathbf$ $|\mathbf a|$ , $|\mathbf b|$ , and $\mathbf a \cdot \mathbf b$ , and then substitute these results into the formula for the angle between these vectors, as shown:

$|\mathbf a|=\sqrt{(1)^2+(-1)^2+(0)^2}=\sqrt{2}$ ,

$|\mathbf b|=\sqrt{(3)^2+(0)^2+(0)^2}=\sqrt{9}=3$ ,

and

$\mathbf a \cdot \mathbf b = (3)(1)+(-1)(0)+(0)(0)=3$ .

Hence,

$\cos\Theta=\frac{3}{3\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$

The principal angle $\Theta$ for which $\cos\Theta=\frac{\sqrt{2}}{2}$ is $\Theta=45^\circ$ . Hence, the angle between the vectors $\mathbf a = \left \langle 1,-1,0\right \rangle$ and $\mathbf b = \left \langle 3,0,0\right \rangle$ measures $45^\circ$ .

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Example Question #55 : Vectors And Vector Operations

Find the angle between the gradient vector $\vec{\triangledown}f(0,y,z)$ and the vector $\vec{u}=4\hat{i}-3\hat{j}+\sqrt{3}\hat{k}$ where is defined as:

$f(x,y,z)=\sqrt{x+1}+y+z$

Possible Answers:

$\theta = \cos^{-1}\left(\frac{1}{3}(\sqrt{7}+\sqrt{3}) \right )$

$\theta = \cos^{-1}\left(\frac{\sqrt{7}}{3} \right )$

$\theta = \cos^{-1}\left(\frac{\sqrt{21}}{3} \right )$

$\theta = \cos^{-1}\left(\sqrt{21} \right )$

$\theta = \cos^{-1}\left(\frac{1}{3}(\sqrt{21}-\sqrt{7}) \right )$

Correct answer:

$\theta = \cos^{-1}\left(\frac{1}{3}(\sqrt{21}-\sqrt{7}) \right )$

Explanation:

Find the angle between the gradient vector $\vec{\triangledown}f(0,y,z)$ and the vector $\vec{u}=4\hat{i}-3\hat{j}+\sqrt{3}\hat{k}$ where is defined as:

$f(x,y,z)=\sqrt{x+1}+y+z$

_____________________________________________________________

Compute the gradient by taking the partial derivative for each direction:

$\vec{\triangledown}f(x,y,z)=\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k}$

$\vec{\triangledown}f(x,y,z)=\frac{1}{2\sqrt{x+1}}\hat{i}+\hat{j}+\hat{k}$

At x = 0 we have:

$\vec{\triangledown}f(0,y,z)=\frac{1}{2}\hat{i}+\hat{j}+\hat{k}$

____________________________________________________________

The angle $\theta$ between two vectors $\vec{v_1}$ and $\vec{v_2}$ can be found using the dot product:

$\vec{v_1}\cdot\vec{v_2}=\left \| \vec{v_1}\right \|\left \| \vec{v_2}\right \|cos(\theta )$

We wish to find the angle $\theta$ between the two vectors:

$\vec{u}=4\hat{i}-3\hat{j}+\sqrt{3}\hat{k}$

$\vec{\triangledown}f(0,y,z)=\frac{1}{2}\hat{i}+\hat{j}+\hat{k}$

Compute the dot product between $\vec{u}$ and $\vec{\triangledown}f(0,y,z)$ ,

$\vec{\triangledown}f(0,y,z)\cdot\vec{u}=\left( \frac{1}{2}\hat{i}+\hat{j}+\hat{k} \right )\cdot\left( 4\hat{i}-3\hat{j}+\sqrt{3}\hat{k}\right )=-1+\sqrt{3}$

Therefore the dot product is:

$\vec{\triangledown}f(0,y,z)\cdot\vec{u}=\sqrt{3}-1$

Compute the magnitude of $\vec{u}$

$\left \| \vec{u}\right \|=\sqrt{16+9+3}=\sqrt{28}=2\sqrt{7}$

Compute the magnitude of $\vec{\triangledown}f(0,y,z)$ ,

$\left \| \vec{\triangledown}f(0,x,y)\right \|=\sqrt{1/4 +2}=\sqrt\frac{9}{4}=\frac{3}{2}$

Now put it all together:

$\vec{\triangledown}f(0,y,z)\cdot\vec{u} =\left \| \vec{\triangledown}f(0,x,y)\right \|\left \| \vec{u}\right \|cos(\theta )$

$\sqrt{3}-1 =3\sqrt{7}\cos(\theta)$

$\theta = \cos^{-1}\left(\frac{\sqrt{21}-\sqrt{7}}{3} \right )$

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Example Question #1 : Distance Between Vectors

Given the vectors

v=(1,0,5)

w=(0,2,4)

find the distance between them, d(v,w) .

Possible Answers:

$\sqrt{5}$

$\sqrt{7}$

$\sqrt{6}$

Correct answer:

$\sqrt{6}$

Explanation:

To find the distance d(v,w) between the vectors

v=(1,0,5)

w=(0,2,4)

we do the following calculation:

$d(v,w)=\sqrt{(1-0)^2+(0-2)^2+(5-4)^2}=\sqrt{1^2+(-2)^2+1^2}$

$=\sqrt{1+4+1}=\sqrt{6}$

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Example Question #1 : Distance Between Vectors

Find the distance d(v,w) between the two vectors

v=(1,3)

w=(5,2)

Possible Answers:

$\sqrt{18}$

$\sqrt{17}$

Correct answer:

$\sqrt{17}$

Explanation:

To find the distance d(v,w) between the two vectors

v=(1,3)

w=(5,2)

we make the following calculation

$d(v,w)=\sqrt{(1-5)^2+(3-2)^2}=\sqrt{16+1}=\sqrt{17}$

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Example Question #1 : Distance Between Vectors

Find the Euclidian distance between the two vectors:

$\mathbf{u}=[5, 3, 0], \, \;\mathbf{v}=[2, -2, \sqrt{2}]$

Possible Answers:

$\frac{9\sqrt{2}}{2}$

$3\sqrt{2}$

Correct answer:

Explanation:

The Euclidian distance between two vectors is:

$d=\sqrt{(u_x-v_x)^2+(u_y-v_y)^2+(u_z-v_z)^2}$

Plugging in the numbers given, we have:

$d=\sqrt{(5-2)^2+(3-(-2))^2+(0-\sqrt2)^2}$

$d=\sqrt{3^2+5^2+\sqrt{2}^2}=\sqrt{9+25+2}=\sqrt{36}=6$

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Example Question #4 : Distance Between Vectors

Find the distance between $\overrightarrow{XY}$ if $X= \left \langle 1,0,-5\right \rangle$ and $Y=\left \langle -3,2,-1\right \rangle$ .

Possible Answers:

$\sqrt{14}$

$3\sqrt2$

$6\sqrt2$

$2\sqrt3$

Correct answer:

Explanation:

Write the formula to find the magnitude of the vector $v=\overrightarrow{XY}$ .

$\left | v\right | = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

Substitute the points into the equation assuming $X=\left \langle x_1,y_1,z_1\right \rangle$ and $Y=\left \langle x_2,y_2,z_2\right \rangle$ .

$\left | v\right | = \sqrt{(-3-1)^2+(2-0)^2+(-1-(-5))^2}$

$\left | v\right | = \sqrt{(-4)^2+(2)^2+4^2}$

$\left | v\right | = \sqrt{16+4+16} = \sqrt{36}= 6$

The answer is:

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Example Question #1 : Distance Between Vectors

Calculate the length of line segment AB given A(−5, −2, 0) and B(6, 0, 3):

Possible Answers:

$\sqrt{134}$

$\sqrt{114}$

$\sqrt{14}$

None of the Above

Correct answer:

$\sqrt{134}$

Explanation:

First we need to find $\vec{AB}$

$\vec{AB}=(6-(-5),0-(-2),3-0)=(11,2,3)$

$\left \| \vec{AB} \right \|= \sqrt{121+4+9}=\sqrt{134}$

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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 #51 : Vectors And Vector Operations

Example Question #51 : Vectors And Vector Operations

Example Question #61 : Calculus 3

Example Question #53 : Vectors And Vector Operations

Example Question #55 : Vectors And Vector Operations

Example Question #1 : Distance Between Vectors

Example Question #1 : Distance Between Vectors

Example Question #1 : Distance Between Vectors

Example Question #4 : Distance Between Vectors

Example Question #1 : Distance Between Vectors

All Calculus 3 Resources

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