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Calculus 3 : Angle between Vectors

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Example Question #1 :Angle Between Vectors

Find the angle between these two vectors, u=(3,4) , and v=(9,-2) .

Possible Answers:

$65.66^\circ$

$65.5^\circ$

$66^\circ$

$50^\circ$

$65.7^\circ$

Correct answer:

$65.66^\circ$

Explanation:

Lets remember the formula for finding the angle between two vectors.

$\cos(\theta)=\frac{(u\cdot v)}{\left \| u\right \|\left \| v\right \|}$

$\cos(\theta)=\frac{((3\cdot 9)+(4\cdot (-2)))}{\sqrt{3^2+4^2}\sqrt{9^2+(-2)^2}}$

$\cos(\theta)=\frac{19}{\sqrt{25}\sqrt{85}}$

$\cos(\theta)=\frac{19}{5\sqrt{85}}$

$\theta=\cos^{-1}\Big(\frac{19}{5\sqrt{85}}\Big)\approx65.66^\circ$

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Example Question #2 :Angle Between Vectors

Calculate the angle between u=(1,2) , v=(3,4) .

Possible Answers:

$10.4^\circ$

$10.3^\circ$

$11^\circ$

$10.5^\circ$

$10^\circ$

Correct answer:

$10.3^\circ$

Explanation:

Lets recall the equation for finding the angle between vectors.

$\cos(\theta)=\frac{(u\cdot v)}{\left \| u\right \|\left \| v\right \|}$

$\cos(\theta)=\frac{((1\cdot 3)+(2\cdot 4))}{\sqrt{1^2+2^2}\sqrt{3^2+4^2}}$

$\cos(\theta)=\frac{11}{\sqrt{5}\sqrt{25}}$

$\cos(\theta)=\frac{11}{5\sqrt{5}}$

$\theta=\cos^{-1}\Big(\frac{11}{5\sqrt{5}}\Big)\approx 10.3^\circ$

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Example Question #3 :Angle Between Vectors

What is the angle between the vectors $\vec{a}= \hat{i}-3\hat{j} +7\hat{k}$ and $\vec{b}= -2\hat{i}+\hat{j} +4\hat{k}$ ?

Possible Answers:

$\theta\approx37.2^{\circ}$

$\theta\approx49.2^{\circ}$

$\theta\approx78.1^{\circ}$

$\theta\approx9.1^{\circ}$

Correct answer:

$\theta\approx49.2^{\circ}$

Explanation:

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

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

where $a\cdot b$ is the dot product of the vectorsand, respectively.

$\left | a\right |$ and $\left | b\right |$ are the magnitudes of vectorsand, respectively.

$cos\theta$ is the angle between the two vectors.

Let vectorbe represented as $\vec{a}= a_{1}\hat{i}+a_{2}\hat{j} +a_{3}\hat{k}$ and vectorbe represented as $\vec{b}= b_{1}\hat{i}+b_{2}\hat{j} +b_{3}\hat{k}$ .

The dot product of the vectorsandis $a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})$ .

The magnitude of vectoris $\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$ and vectoris $\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}$ .

Rearranging the dot product formula to solve for $\theta$ gives us $\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )$

For this problem,

$a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})=1(-2)+(-3)(1)+(7)(4)$

$a\cdot b=-2-3+28=23$

$\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}=\sqrt{(1)^{2}+(-3)^{2}+(7)^{2}}$

$\left | a\right |=\sqrt{1+9+49}=\sqrt{59}$

$\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}=\sqrt{(-2)^{2}+(1)^{2}+(4)^{2}}$

$\left | b\right |=\sqrt{4+1+16}=\sqrt{21}$

$\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )=cos^{-1}\left ( \frac{23}{\sqrt{59}\sqrt{21}}\right )=cos^{-1}\left ( \frac{23}{\sqrt{1239}}\right )$

$\theta\approx49.2^{\circ}$

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Example Question #4 :Angle Between Vectors

What is the angle between the vectors $\vec{a}=2 \hat{i}-1\hat{j} +7\hat{k}$ and $\vec{b}= \hat{i}+2\hat{j}$ ?

Possible Answers:

$\theta\approx45^{\circ}$

$\theta\approx8.1^{\circ}$

$\theta\approx90^{\circ}$

$\theta\approx0^{\circ}$

Correct answer:

$\theta\approx90^{\circ}$

Explanation:

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

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

where $a\cdot b$ is the dot product of the vectorsand, respectively.

$\left | a\right |$ and $\left | b\right |$ are the magnitudes of vectorsand, respectively.

$cos\theta$ is the angle between the two vectors.

Let vectorbe represented as $\vec{a}= a_{1}\hat{i}+a_{2}\hat{j} +a_{3}\hat{k}$ and vectorbe represented as $\vec{b}= b_{1}\hat{i}+b_{2}\hat{j} +b_{3}\hat{k}$ .

The dot product of the vectorsandis $a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})$ .

The magnitude of vectoris $\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$ and vectoris $\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}$ .

Rearranging the dot product formula to solve for $\theta$ gives us $\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )$

For this problem,

$a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})=2(1)+(-1)(2)+(7)(0)$

$a\cdot b=2-2+0=0$

$\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}=\sqrt{(2)^{2}+(1)^{2}+(7)^{2}}$

$\left | a\right |=\sqrt{4+1+49}=\sqrt{54}$

$\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}=\sqrt{(1)^{2}+(2)^{2}+(0)^{2}}$

$\left | b\right |=\sqrt{1+4}=\sqrt{5}$

$\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )=cos^{-1}\left ( \frac{0}{\sqrt{54}\sqrt{5}}\right )=cos^{-1}\left (0\right )$

$\theta\approx90^{\circ}$

The vectors are perpendicular

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Example Question #5 :Angle Between Vectors

What is the angle between the vectors $\vec{a}=-2 \hat{i}-3\hat{j} +\hat{k}$ and $\vec{b}= \hat{i}+2\hat{j}+5\hat{k}$ ?

Possible Answers:

$\theta\approx8.37^{\circ}$

$\theta\approx81.58^{\circ}$

$\theta\approx120.8^{\circ}$

$\theta\approx98.42^{\circ}$

Correct answer:

$\theta\approx98.42^{\circ}$

Explanation:

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

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

where $a\cdot b$ is the dot product of the vectorsand, respectively.

$\left | a\right |$ and $\left | b\right |$ are the magnitudes of vectorsand, respectively.

$cos\theta$ is the angle between the two vectors.

Let vectorbe represented as $\vec{a}= a_{1}\hat{i}+a_{2}\hat{j} +a_{3}\hat{k}$ and vectorbe represented as $\vec{b}= b_{1}\hat{i}+b_{2}\hat{j} +b_{3}\hat{k}$ .

The dot product of the vectorsandis $a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})$ .

The magnitude of vectoris $\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$ and vectoris $\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}$ .

Rearranging the dot product formula to solve for $\theta$ gives us $\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )$

For this problem,

$a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})=-2(1)+(-3)(2)+(1)(5)$

$a\cdot b=-2-6+5=-3$

$\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}=\sqrt{(-2)^{2}+(-3)^{2}+(1)^{2}}$

$\left | a\right |=\sqrt{4+9+1}=\sqrt{14}$

$\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}=\sqrt{(1)^{2}+(2)^{2}+(5)^{2}}$

$\left | b\right |=\sqrt{1+4+25}=\sqrt{30}$

$\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )=cos^{-1}\left ( \frac{-3}{\sqrt{14}\sqrt{30}}\right )=cos^{-1}\left (\frac{-3}{\sqrt{420}}\right )$

$\theta\approx98.42^{\circ}$

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Example Question #6 :Angle Between Vectors

What is the angle between the vectors $\vec{a}= \hat{i}+3\hat{j} +5\hat{k}$ and $\vec{b}= 2\hat{i}+6\hat{j}+10\hat{k}$ ?

Possible Answers:

$\theta\approx90^{\circ}$

$\theta\approx8.37^{\circ}$

$\theta\approx0^{\circ}$

$\theta\approx120^{\circ}$

Correct answer:

$\theta\approx0^{\circ}$

Explanation:

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

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

where $a\cdot b$ is the dot product of the vectorsand, respectively.

$\left | a\right |$ and $\left | b\right |$ are the magnitudes of vectorsand, respectively.

$cos\theta$ is the angle between the two vectors.

Let vectorbe represented as $\vec{a}= a_{1}\hat{i}+a_{2}\hat{j} +a_{3}\hat{k}$ and vectorbe represented as $\vec{b}= b_{1}\hat{i}+b_{2}\hat{j} +b_{3}\hat{k}$ .

The dot product of the vectorsandis $a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})$ .

The magnitude of vectoris $\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$ and vectoris $\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}$ .

Rearranging the dot product formula to solve for $\theta$ gives us $\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )$

For this problem,

$a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})=1(2)+(3)(6)+(5)(10)$

$a\cdot b=2+18+50=70$

$\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}=\sqrt{(1)^{2}+(3)^{2}+(5)^{2}}$

$\left | a\right |=\sqrt{1+9+25}=\sqrt{35}$

$\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}=\sqrt{(2)^{2}+(6)^{2}+(10)^{2}}$

$\left | b\right |=\sqrt{4+36+100}=\sqrt{140}$ $\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )=cos^{-1}\left ( \frac{70}{\sqrt{35}\sqrt{140}}\right )=cos^{-1}\left (\frac{70}{\sqrt{4900}}\right )$

$\theta=cos^{-1}\left (\frac{70}{70}\right )=cos^{-1}\left (1\right )$

$\theta\approx0^{\circ}$

The two vectors are parallel.

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Example Question #7 :Angle Between Vectors

Find the approximate acute angle in degrees between the vectors <2,2,5>, <0,5,-1> .

Possible Answers:

29.77

80.17

79.23

12.55

None of the other answers

Correct answer:

80.17

Explanation:

To find the angle between two vectors, use the formula

$\theta = \cos^{-1}(\frac{\mathbf{a}\cdot\mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|})$ .

$\theta = \cos^{-1}(\frac{<2,2,5>\cdot<0,5,-1>}{\sqrt{4+4+25}\sqrt{0+25+1}})$

$\theta = \cos^{-1}(\frac{5}{\sqrt{858}})$

$\theta \approx 80.22$

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Example Question #8 :Angle Between Vectors

Find the angle between the following two vectors.

$\vec{A}=3\hat{i}+5\hat{j}-\hat{k}, \vec{B}=6\hat{i}-2\hat{j}+5\hat{k}$

Possible Answers:

$86.39 ^{\circ}$

$18.41 ^{\circ}$

$32.57 ^{\circ}$

$3.61 ^{\circ}$

Correct answer:

$86.39 ^{\circ}$

Explanation:

In order to find the angle between two vectors, we need to take the quotient of their dot product and their magnitudes:

$\frac{\vec{A}\cdot \vec{B}}{\left|\vec{A} \right| \left| \vec{B} \right| } = \cos{\theta}\Rightarrow \theta = \cos^{-1} \left( \frac{\vec{A}\cdot \vec{B}}{\left|\vec{A} \right| \left| \vec{B} \right| }\right )$

Therefore, we find that

$\vec{A}\cdot\vec{B}=3(6)+5(-2)-1(5)=3,$

$\left| \vec{A}\right| = \sqrt{3^2+5^2+1^2}=\sqrt{35}, \left| \vec{B}\right| = \sqrt{6^2+2^2+5^2}=\sqrt{65}$

$\theta = \cos^{-1} \left( \frac{3}{\sqrt{35}\sqrt{65} }\right)=86.39 ^{\circ}$ .

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Example Question #9 :Angle Between Vectors

Find the (acute) angle between the vectors $<1,0,\frac{1}{2}>, <\frac{3}{2},0, 12>$ in degrees.

Possible Answers:

8.659

82.341

42.690

40.542

56.310

Correct answer:

56.310

Explanation:

To find the angle between vectors, we use the formula

$\theta = \cos^{-1}(\frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|})$ .

Substituting in our values, we get

$\theta = \cos^{-1}(\frac{<1,0,\frac{1}{2}> \cdot <\frac{3}{2},0,12>}{\sqrt{1^2+0^2+\frac{1}{2}^2}\sqrt{\frac{3}{2}^2+0^2+12^2}}) \approx \cos^{-1}(\frac{7.5}{13.521}) = 56.310$

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Example Question #10 :Angle Between Vectors

Find the angle between the two vectors.

u=<2,0>
$v=<\sqrt{3}, 1>$

Possible Answers:

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

No angle exists

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

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

Correct answer:

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

Explanation:

To find the angle between two vector we use the following formula

$cos \, \theta = \frac{u\bullet v}{\left \| u\right \|*\left \| v\right \|}$

and solve for $\theta$ .

鉴于

u=<2,0>
$v=<\sqrt{3}, 1>$ we find

$u\bullet v= 2*\sqrt3+0*1=2\sqrt3$

$\left \| u\right \|=\sqrt{2^2+0^2}=\sqrt4=2$

$\left \| v\right \|=\sqrt{\sqrt{3}^2+1^2}=\sqrt4=2$

Plugging these values in we get

$cos\, \theta = \frac{2\sqrt3}{2*2}=\frac{\sqrt3}{2}$

To find $\theta$ we calculate the $cos^{-1}$ of both sides

$cos^{-1}[cos(\theta)]=cos^{-1}(\frac{\sqrt3}{2})$

and find that

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

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Calculus 3 : Angle between Vectors

Study concepts, example questions & explanations for Calculus 3

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Example Question #1 :Angle Between Vectors

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