AP Calculus BC - Angle between Vectors

Find the angle between the two vectors

$\theta=58^o$

$\theta=25^o$

$\theta=70^o$

$\theta=90^o$

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

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

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

$\theta=58^o$

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

None of the other answers

Explanation

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.)

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

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

1.46

1.18

0.98

2.89

1.62

Explanation

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:

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

Find the angle between the two vectors.

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

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

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

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

No angle exists

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 $\theta$ .

Given

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

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

Find the angle between vectors $a=\left \langle 1,0,3\right \rangle$ and $b=\left \langle 2,1,1\right \rangle$ . Use the dot product when finding the solution.

$\cos^{-1}(\frac{5}{\sqrt{60}})$

$\cos^{-1}(\frac{2}{\sqrt5})$

$\cos^{-1}({\frac{4}{9}})$

$\sin^{-1}(\frac{5}{\sqrt60})$

Explanation

First, we must find the magnitude of the vectors.

$\left | a\right |=\sqrt{1^2+3^2}=\sqrt{10}$

$\left | b\right |=\sqrt{2^2+1^2+1^2}=\sqrt{6}$

Next, we find the dot product

$a\cdot b=(2*1)+(0*1)+(3*1)=5$

Plugging into the dot product formula, we get

$5=\sqrt{10*6}\cos\theta$ . Solving for theta, we then get

$\theta=\cos^{-1}(\frac{5}{\sqrt{60}})$

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

$\theta=118^o$

$\theta=67^o$

$\theta=143^o$

$\theta=35^o$

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

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

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

$\theta=118^o$

Find the angle $\theta$ to the nearest degree between the two vectors

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

$cos(\theta)=\frac{<4,14>\bullet <-3,-2>}{\sqrt{4^2+14^2}*\sqrt{(-3)^2+(-2)^2}}=\frac{-12+(-28)}{\sqrt{260}*\sqrt{13}}$

Simplifying we get

$cos(\theta)=\frac{-40}{\sqrt{3380}}$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{-40}{\sqrt{3380}})$

and find that

$\theta=133^o$

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.

$\cos^{-1}(\frac{2}{\sqrt{10}})$

$\cos^{-1}(\frac{2}{10})$

$\sin^{-1}(\frac{1}{5})$

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 $\theta=\cos^{-1}(\frac{2}{{\sqrt{10}}})$ .

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 $a\cdot b=11$ .

$\theta=\cos^{-1}(\frac{11}{\sqrt{273}})$

$\theta=\cos^{-1}(\frac{6}{\sqrt{73}})$

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

$\theta=\cos^{-1}(\frac{10}{4})$

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

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 $\theta=\cos^{-1}(\frac{11}{\sqrt{273}})$