Angle between Vectors - AP Calculus BC

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Question

Find the angle between these two vectors, , and .

Answer

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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Question

Calculate the angle between , .

Answer

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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Question

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

Answer

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 vectors and , respectively.

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

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

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

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

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

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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Question

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

Answer

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 vectors and , respectively.

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

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

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

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

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

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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Question

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

Answer

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 vectors and , respectively.

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

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

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

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

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

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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Question

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

Answer

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 vectors and , respectively.

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

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

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

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

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

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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Question

Find the angle between the two vectors.

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

Answer

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 $\theta$ .

Given

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

$u\bullet v= 2\sqrt3+01=2\sqrt3$

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

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

Plugging these values in we get

$cos, \theta = \frac{2\sqrt3}{22}=\frac{\sqrt3}{2}$

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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Question

Find the approximate acute angle in degrees between the vectors .

Answer

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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Question

Find the angle between the following two vectors.

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

Answer

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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Question

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

Answer

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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Question

Find the angle between the two vectors.

Answer

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 $\theta$ .

Given

we find

$u\bullet v= 1(-3)+13=0$

$\left | u\right |=\sqrt{1^2+1^2}=\sqrt2$

$\left | v\right |=\sqrt{(-3)^2+3^2}=\sqrt{18}=3\sqrt2$

Plugging these values in we get

$cos, \theta = \frac{0}{3\sqrt2\sqrt2}=0$

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

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

and find that

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

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Question

Find the approximate angle in degrees between the vectors $<1, e, e^2>, <e, \sqrt{e}, \sqrt[3]{e}>$ .

Answer

We can compute the (acute) angle between the two vectors using the formula

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

Hence we have

$\theta = \cos^{-1}(\frac{<1,e,e^2>\cdot<e,\sqrt{e},\sqrt[3]{e}>}{\sqrt{1+e^2+e^4}\sqrt{e^2+e+e^{3/2}}})$

$\approx \cos^{-1}(\frac{e+e\sqrt{e}+e^{7/3}}{30.314})$

$\approx \cos^{-1}(\frac{17.512}{30.314})$

$\approx 54.712$

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Question

Find the angle in degrees between the vectors $<1,1>, <\pi^e,e^{\pi}>$ .

Answer

The correct answer is approximately degrees.

To find the angle between two vectors, we use the equation $\theta = \cos^{-1}(\frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|})$ .

Hence we have

$\theta = \cos^{-1}(\frac{(1)(\pi^e)+(1)(e^{\pi})}{\sqrt{1^2+1^2}\sqrt{\pi^{2e}+e^{2\pi}}})$

(This answer is small due to the fact that the two vectors nearly point in the same direction, due to $e^\pi$ and $\pi^e$ being close in value.)

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Question

Find the acute angle in degrees between the vectors .

Answer

To find the angle between two vectors, we use the formula

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

So we have

$\theta = \cos^{-1}(\frac{<10,1,5>\cdot<1,-1,-1>}{\sqrt{10^2+1^2+5^2}\sqrt{1^2+(-1)^2+(-1)^2}})$

$\theta = \cos^{-1}(\frac{4}{\sqrt{126}\sqrt{3}})$

$\theta \approx 78.127$

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Question

Find the angle between the following vectors (to two decimal places):

$\mathbf{v}=[-2, 4, -1],;\mathbf{w}=[3, 1, -5]$

Answer

The dot product is defined as:

$\mathbf{v}\cdot \mathbf{w}=|\mathbf{v}||\mathbf{w}|\cos\theta$

Where theta is the angle between the two vectors. Solving for theta:

$\theta=\cos^{-1}(\frac{\mathbf{v}\cdot \mathbf{w}}{|\mathbf{v}||\mathbf{w}|})$

To solve each component:

$\mathbf{v}\cdot \mathbf{w}=(3)(-2)+(1)(4)+(-5)(-1)=-6+4+5=3$

$|\mathbf{v}|=\sqrt{3^2+1^2+(-5)^2}=\sqrt{9+1+25}=\sqrt{35}$

$|\mathbf{w}|=\sqrt{(-2)^2+4^2+(-1)^2}=\sqrt{4+16+1}=\sqrt{21}$

Putting it all together, we can solve for theta:

$\theta=\cos^{-1}(\frac{3}{\sqrt{35}\sqrt{21}})\approx 1.46$

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Question

Find the angle between vectors $u= \left \langle 1,-2,3\right \rangle$ and $v=\left \langle -3,-4,-6\right \rangle$ and round to the nearest degree.

Answer

Write the formula to find the angle between two vectors.

$\theta= cos^{-1}(\frac{u_1v_1+u_2v_2+u_3v_3}{\left | u\right |\left | v\right |})$

Evaluate each term.

$\left | u\right | = \sqrt{1^2+(-2)^2+(3)^2} = \sqrt{1+4+9} = \sqrt{14}$

$\left | v\right | = \sqrt{(-3)^2+(-4)^2+(-6)^2} = \sqrt{9+16+36}= \sqrt{61}$

$\left | u\right |\left | v\right | = \sqrt{14}\sqrt{61} = \sqrt{ 854}$

Substitute the values into the equation.

$\theta= cos^{-1}(\frac{-13}{\sqrt{854}})= 116.41379$

The correct answer is:

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Question

Find the angle between the two vectors

Answer

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

$cos(\theta)=\frac{<3,7>\bullet <-7,10>}{\sqrt{3^2+7^2}\sqrt{(-7)^2+10^2}}=\frac{3(-7)+7(10)}{\sqrt{58}*\sqrt{149}}$

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

$\theta=58^o$

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Question

What is the angle to the nearest degree between the vectors and ?

Answer

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

$cos(\theta)=\frac{<4,5>\bullet <3,-7>}{\sqrt{4^2+5^2}\sqrt{3^2+(-7)^2}}=\frac{43+5(-7)}{\sqrt{41}\sqrt{58}}$

Simplifying we get

$cos(\theta)=\frac{-23}{\sqrt{2378}}$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{-23}{\sqrt{2378}})$

and find that

$\theta=118^o$

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Question

Find the angle between the two vectors

and

Round to the nearest degree.

Answer

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

$cos(\theta)=\frac{<3,4>\bullet <4,3>}{\sqrt{3^2+4^2}\sqrt{4^2+3^2}}=\frac{34+43}{\sqrt{25}\sqrt{25}}$

Simplifying we get

$cos(\theta)=\frac{24}{25}$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{24}{25})$

and find that

$\theta=16^o$

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Question

Find the angle between the two vectors

Answer

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

$cos(\theta)=\frac{<5,0>\bullet <-3,-3>}{\sqrt{5^2+0^2}\sqrt{(-3)^2+(-3)^2}}=\frac{5(-3)+0(-3)}{\sqrt{25}*\sqrt{18}}$

Simplifying we get

$cos(\theta)=\frac{-15}{\sqrt{450}}$

To solve for $\theta$ we find the $cos^{-1}$ of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{-15}{\sqrt{450}})$

and find that

$\theta=135^o$

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