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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 #401 : Vectors And Vector Operations

Find the dot product of the two vectors

u=<3,0>

v=<-5,0>

Possible Answers:

Correct answer:

Explanation:

The dot product for the vectors

a=<a_1,a_2>

b=<b_1,b_2>

is defined as

$a \bullet b = a_1*b_1+a_2*b_2$

For the vectors in this problem we find that

$u \bullet v=3(-5)+0*0=-15$

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Example Question #402 : Vectors And Vector Operations

Find the dot product of the two vectors

u=<1,0>

v=<0,1>

Possible Answers:

Correct answer:

Explanation:

The dot product for the vectors

a=<a_1,a_2>

b=<b_1,b_2>

is defined as

$a\bullet b=a_1*b_1+a_2*b_2$

For the vectors in this problem we find that

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

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Example Question #411 : Vectors And Vector Operations

Calculate the dot product

$<10,4>\bullet<1,0>$

Possible Answers:

Correct answer:

Explanation:

The dot product for the vectors

a=<a_1,a_2>

b=<b_1,b_2>

is defined as

$a\bullet b=a_1*b_1+a_2*b_2$

For the vectors in this problem we find that

$<10,4>\bullet<1,0>=10*1+4*0=10$

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Example Question #2411 : Calculus 3

Find the vector projection of <1,1> onto <-1,2> .

Possible Answers:

<2,-1>

<-1,2>

$<-\frac{1}{3}, \frac{2}{3}>$

None of the other answers

$<-\frac{3}{2}, 1>$

Correct answer:

$<-\frac{1}{3}, \frac{2}{3}>$

Explanation:

To find the vector projection of $\mathbf{b}$ onto $\mathbf{a}$ , we take the scalar projection $\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|}$ , and multiply it to a vector of unit length; $\frac{\mathbf{a}}{|\mathbf{a}|}$ .

Hence for the vector projection of <1,1> onto <-1,2> , we have

$\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|}$ $\frac{\mathbf{a}}{|\mathbf{a}|}$ $= \frac{<-1,2> \cdot <1,1>}{\sqrt{(-1)^2+2^2}}$ $\frac{<-1,2>}{\sqrt{(-1)^2+2^2}}$

$=\frac{1}{\sqrt{3}}$ $\frac{1}{\sqrt{3}}<-1,2>$ $=<-\frac{1}{3},\frac{2}{3}>$ .

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Example Question #91 : Dot Product

Find the vector projection of <-1,1,0> onto <0,1,1> .

Possible Answers:

$<0,\frac{1}{2},-1>$

None of the other answers.

<1,-1,0>

$<-\frac{1}{2},\frac{1}{2},0>$

$<-1,0,\frac{1}{2}>$

Correct answer:

$<-\frac{1}{2},\frac{1}{2},0>$

Explanation:

Hence for the vector projection of <-1,1,0> onto <0,1,1> , we have

$\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|}$ $\frac{\mathbf{a}}{|\mathbf{a}|}$ $= \frac{<-1,1,0> \cdot <0,1,1>}{\sqrt{(-1)^2+1^2+0^2}}$ $\frac{<-1,1,0>}{\sqrt{(-1)^2+1^2+0^2}}$

$=\frac{1}{\sqrt{2}}$ $\frac{1}{\sqrt{2}}<-1,1,0>$ $=<-\frac{1}{2},\frac{1}{2},0>$ .

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Example Question #2411 : Calculus 3

Find the dot product of $\vec{A}=2i+3j+k$ and $\vec{B}=3i+4j+12k$

Possible Answers:

Correct answer:

Explanation:

$\vec{A}=\begin{pmatrix} 2\\3 \\1 \end{pmatrix}$ $\vec{B}=\begin{pmatrix} 3\\4 \\12 \end{pmatrix}$

$\vec{A}.\vec{B}=2\times3+3\times4+1\times12=30$

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Example Question #2412 : Calculus 3

Simplify:

$\left \langle x+y, x^2\cos(z), z\right \rangle \cdot \left \langle 2, 2, 3\right \rangle$

Possible Answers:

$\left \langle \2x+2y,2x^2\cos(z),3z \right \rangle$

$2x+2y+2x^2\cos(z)+3z$

$2x+2y+2x^2\cos(z)+2z$

$7x+7y+7x^2\cos(z)+7z$

Correct answer:

$2x+2y+2x^2\cos(z)+3z$

Explanation:

The dot product of two vectors is given by:

$\left \langle a, b\right \rangle \cdot \left \langle c, d\right \rangle = ac+bd$

Using this, for our vectors, we get

$2x+2y+2x^2\cos(z)+3z$

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Example Question #2413 : Calculus 3

Solve:

$\left \langle 0, 1, 1 \right \rangle \cdot \left \langle 3, 2, 5 \right \rangle$

Possible Answers:

Correct answer:

Explanation:

The dot product of two vectors is given by

$\left \langle a, b, c \right \rangle \cdot \left \langle d, e, f \right \rangle= ad+be+cf$

Using this, we get

0+2+5=7

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Example Question #2414 : Calculus 3

Solve:

$\left \langle 1, 1, 2 \right \rangle \cdot \left \langle \cos(2), 0, 2 \right \rangle$

Possible Answers:

$4\cos(2)$

$\cos(2)+5$

$\cos(2)+4$

Correct answer:

$\cos(2)+4$

Explanation:

The dot product of two vectors is given by

$\left \langle a, b,c \right \rangle \cdot \left \langle d, e, f \right \rangle = ad+be+cf$

Using this, we get

$\cos(2)+0+4=\cos(2)+4$

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Example Question #2415 : Calculus 3

Solve:

$\left \langle x, 2xy \right \rangle\cdot \left \langle ze^{y}, 3x^2 \right \rangle$

Possible Answers:

xze^y-6x^3y

xze^y+6x^2y

$xze^{y}+6x^3y$

6x^4yze^y

Correct answer:

$xze^{y}+6x^3y$

Explanation:

The dot product of two vectors is given by

$\left \langle a, b \right \rangle \cdot \left \langle c, d \right \rangle= ac+bd$

Our final answer is

$xze^{y}+6x^3y$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #401 : Vectors And Vector Operations

Example Question #402 : Vectors And Vector Operations

Example Question #411 : Vectors And Vector Operations

Example Question #2411 : Calculus 3

Example Question #91 : Dot Product

Example Question #2411 : Calculus 3

Example Question #2412 : Calculus 3

Example Question #2413 : Calculus 3

Example Question #2414 : Calculus 3

Example Question #2415 : Calculus 3

All Calculus 3 Resources

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