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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #16 : Angle Between Vectors

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.

Possible Answers:

148

132

128

116

Correct answer:

116

Explanation:

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.

u_1v_1 = 1(-3)=-3

u_2v_2 = (-2)(-4)=8

u_3v_3 = 3(-6) = -18

u_1v_1+u_2v_2+u_3v_3= -13

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

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

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

Substitute the values into the equation.

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

The correct answer is: 116

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Example Question #17 : Angle Between Vectors

Find the angle between the two vectors

u=<3,7>

v=<-7,10>

Possible Answers:

$\theta=70^o$

$\theta=58^o$

$\theta=90^o$

$\theta=25^o$

Correct answer:

$\theta=58^o$

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

$\theta=58^o$

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Example Question #18 : Angle Between Vectors

What is the angle to the nearest degree between the vectors <4,5> and <3,-7> ?

Possible Answers:

$\theta=118^o$

$\theta=143^o$

$\theta=67^o$

$\theta=35^o$

Correct answer:

$\theta=118^o$

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

$\theta=118^o$

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Example Question #19 : Angle Between Vectors

Find the angle between the two vectors

<3,4> and <4,3>

Round to the nearest degree.

Possible Answers:

$\theta=10^o$

$\theta=16^o$

$\theta=54^o$

$\theta=30^o$

Correct answer:

$\theta=16^o$

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

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

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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Example Question #20 : Angle Between Vectors

Find the angle between the two vectors

u=<5,0>

v=<-3,-3>

Possible Answers:

$\theta=135^o$

$\theta=90^o$

$\theta=60^o$

$\theta=45^o$

Correct answer:

$\theta=135^o$

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

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

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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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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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #16 : Angle Between Vectors

Example Question #17 : Angle Between Vectors

Example Question #18 : Angle Between Vectors

Example Question #19 : Angle Between Vectors

Example Question #20 : Angle Between Vectors

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

All Calculus 3 Resources

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