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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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

Example Questions

Example Question #26 : Vectors And Vector Operations

Find the angle between the two vectors.

u=<-3,6>

v=<5,7>

Possible Answers:

$\theta=81.3^o$

$\theta=96.6^o$

$\theta=42.5^o$

$\theta=62.1^o$

Correct answer:

$\theta=62.1^o$

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

$cos(\theta)=\frac{-3*5+6*7}{\sqrt{(-3)^2+6^2}*\sqrt{5^2+7^2}}$

Simplifying we get

$cos(\theta)=\frac{27}{\sqrt{3330}}$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{27}{\sqrt{3330}})$

and find that

$\theta=62.1^o$

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Example Question #27 : Vectors And Vector Operations

Find the angle between vectors and .

u=<10,20>

v=<7,10>

Round your answer to the nearest degree.

Possible Answers:

$\theta=102.5^o$

$\theta=34.6^o$

$\theta=57.1^o$

$\theta=8.4^o$

Correct answer:

$\theta=8.4^o$

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

$cos(\theta)=\frac{10*7+20*10}{\sqrt{10^2+20^2}*\sqrt{7^2+10^2}}$

Simplifying we get

$cos(\theta)=\frac{270}{\sqrt{74500}}$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{270}{\sqrt{74500}})$

and find that

$\theta=8.4^o$

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Example Question #28 : Vectors And Vector Operations

Find the angle between vectors a and b, where $\left | a\right |=4$ , $\left | b\right |=\frac{1}{2}$ , and $a\cdot b=\sqrt2$ .

Possible Answers:

$\pi$

$\frac{\pi}{2}$

$2\pi$

$\frac{\pi}{4}$

Correct answer:

$\frac{\pi}{4}$

Explanation:

Using the formula for the dot product, which is

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

all we do not have is the value of theta.

Plugging in what we know, we get

$\cos\theta=\frac{\sqrt2}{2}$ .

Doing inverse cosine of both sides gets us

$\theta=\frac{\pi}{4}$ .

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Example Question #29 : Vectors And Vector Operations

Find the angle between the vectors a and b, where $\left | a\right |=4$ , $\left | b\right |=\frac{1}{2}$ , and $a\times b=\sqrt3$ .

Possible Answers:

$\pi$

$\frac{\pi}{4}$

$\frac{\pi}{2}$

$\frac{\pi}{3}$

Correct answer:

$\frac{\pi}{3}$

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

$\sin(\theta)=\frac{\sqrt3}{2}$ .

Therefore,

$\theta=\sin^{-1}(\frac{\sqrt3}{2})=\frac{\pi}{3}$

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Example Question #30 : Vectors And Vector Operations

Find the angle between the vectors a and b, where $\left | a\right |=4$ , $\left | b\right |=\frac{1}{2}$ , and $a\cdot b=\sqrt3$ .

Possible Answers:

$\frac{\pi}{6}$

$\pi$

$\frac{\pi}{3}$

Correct answer:

$\frac{\pi}{6}$

Explanation:

Using the formula for the dot product, which is

$a\cdot b=\left | a\right |\left | b\right |\cos(\theta)$ ,

we have all the values except theta.

Plugging in the known values and solving, we get

$\cos(\theta)=\frac{\sqrt3}{2}$ .

Therefore,

$\theta=\cos^{-1}(\frac{\sqrt3}{2})=\frac{\pi}{6}$

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Example Question #31 : Vectors And Vector Operations

Find the direction angles of the vector $i-j+\sqrt{2}k$

Possible Answers:

$\alpha=\frac{\pi}{3}, \beta=\frac{2\pi}{3}, \gamma=\frac{\pi}{4}$

None of the Above

$\alpha=\frac{\pi}{3}, \beta=\frac{\pi}{3}, \gamma=\frac{3\pi}{4}$

$\alpha=\frac{2\pi}{3}, \beta=-\frac{\pi}{3}, \gamma=\frac{3\pi}{4}$

$\alpha=\frac{\pi}{3}, \beta=\frac{\pi}{3}, \gamma=\frac{\pi}{4}$

Correct answer:

$\alpha=\frac{\pi}{3}, \beta=\frac{2\pi}{3}, \gamma=\frac{\pi}{4}$

Explanation:

To find the direction angles we must first find the Unit vector of $i-j+\sqrt{2}k$ .

$\left \| (1,-1,\sqrt{2}) \right \|=\sqrt{1+1+2}=\sqrt{4}=2$

$\vec{u}=(\frac{1}{2},-\frac{1}{2},\frac{\sqrt{2}}{2})$

Then we use Cosine to find each angle:

$\cos \alpha=\frac{1}{2}, \cos \beta=-\frac{1}{2}, \cos \gamma=\frac{\sqrt{2}}{2}$

so,

$\alpha=\frac{\pi}{3}, \beta=\frac{2\pi}{3}, \gamma=\frac{\pi}{4}$

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Example Question #32 : Vectors And Vector Operations

If a = (3,2,−1) and b = (6,α,−2) are parallel, then α =

Possible Answers:

None of the Above

Correct answer:

Explanation:

If a and b are parallel, then there is a scaler multiple of $\vec{b}=\beta \vec{a}$ :

in this case $\beta=2$ . Therefore,

$\vec{b}=2\vec{a}$

so,

$\alpha=2*2=4$

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Example Question #33 : Vectors And Vector Operations

Find the angle between the vectors

u=<3,17>

v=<5,2>

Round to the nearest tenth.

Possible Answers:

$\theta=34.1^o$

$\theta=81.3^o$

$\theta=104.9^o$

$\theta=58.2^o$

Correct answer:

$\theta=58.2^o$

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

$cos(\theta)=\frac{3(5)+17(2)}{\sqrt{3^2+17^2}*\sqrt{5^2+2^2}}$

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

$\theta=58.2^o$

Report an Error

Example Question #31 : Vectors And Vector Operations

Find the angle between the vectors and , given that $\left | a\right |=1$ , $\left | b\right |=2$ , and $a\cdot b=\sqrt2$ .

Possible Answers:

$\theta=\frac{\pi}{4}$

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

$\theta=2\pi$

$\theta=\pi$

Correct answer:

$\theta=\frac{\pi}{4}$

Explanation:

Using the dot product formula $a\cdot b=\left | a\right |\left | b\right |\cos(\theta)$ . Plugging in what we were given in the problem statement, we get $\sqrt2=2\cos(\theta)$ . Solving for $\theta$ we get $\cos^{-1}(\frac{\sqrt2}{2})=\theta=\frac{\pi}{4}$ .

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Example Question #35 : Vectors And Vector Operations

Find the angle between the vectors and , given that $\left | a\right |=1$ , $\left | b\right |=1$ , and $a\cdot b= -1$ .

Possible Answers:

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

$\theta=\frac{\pi}{4}$

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

$\theta=\pi$

Correct answer:

$\theta=\pi$

Explanation:

Using the dot product formula $a\cdot b=\left | a\right |\left | b\right |\cos(\theta)$ . Plugging in what we were given in the problem statement, we get $-1=1\cos(\theta)$ . Solving for $\theta$ we get $\cos^{-1}(-1)=\theta=\pi$ .

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #26 : Vectors And Vector Operations

Example Question #27 : Vectors And Vector Operations

Example Question #28 : Vectors And Vector Operations

Example Question #29 : Vectors And Vector Operations

Example Question #30 : Vectors And Vector Operations

Example Question #31 : Vectors And Vector Operations

Example Question #32 : Vectors And Vector Operations

Example Question #33 : Vectors And Vector Operations

In order to find the angle between the two vectors, we follow the formula

Example Question #31 : Vectors And Vector Operations

Example Question #35 : Vectors And Vector Operations

All Calculus 3 Resources

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