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Calculus 3 : Dot Product

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Example Questions

Example Question #91 : Dot Product

Find the length $\small \small ||v||$ of the vector $\small \small \small \small \small \small \small \small \small \small v=(1,-1,2)$ .

Possible Answers:

$\small \small \small \small ||v||=4$

$\small \small \small ||v||=\sqrt{6}$

$\small \small \small \small ||v||=7$

$\small \small \small \small ||v||=\sqrt{18}$

Correct answer:

$\small \small \small ||v||=\sqrt{6}$

Explanation:

To find the length $\small \small ||v||$ of the vector $\small \small \small \small \small \small \small \small \small \small v=(1,-1,2)$ , we take the square root of the dot product $\small v\cdot v$ :

$\small \small \small \small \small \small |v||=\sqrt{v\cdot v}=\sqrt{1^2+(-1)^2+2^2}=\sqrt{1+1+4}=\sqrt{6}$

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Example Question #92 : Dot Product

Find the length $\small \small ||v||$ of the vector $\small \small \small \small \small \small \small \small \small \small \small v=(\sqrt{5},0,\sqrt{8})$ .

Possible Answers:

$\small \small \small \small ||v||=13$

$\small \small \small \small ||v||=\frac{1}{2}(\sqrt{5}+\sqrt{8})$

$\small \small \small \small ||v||=\sqrt{5}+\sqrt{8}$

$\small \small \small ||v||=\sqrt{13}$

Correct answer:

$\small \small \small ||v||=\sqrt{13}$

Explanation:

To find the length $\small \small ||v||$ of the vector $\small \small \small \small \small \small \small \small \small \small \small v=(\sqrt{5},0,\sqrt{8})$ , we take the square root of the dot product $\small v\cdot v$ :

$\small \small \small \small \small \small \small |v||=\sqrt{v\cdot v}=\sqrt{\sqrt{5}^2+0^2+\sqrt{8}^2}=\sqrt{5+8}=\sqrt{13}$

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Example Question #93 : Dot Product

Find the length $\small \small ||v||$ of the vector $\small \small \small v=(-\sqrt{5},\sqrt{20})$ .

Possible Answers:

$\small \small \small \small ||v||=\sqrt{20}-\sqrt{5}$

$\small \small \small \small ||v||=\sqrt{5}+\sqrt{20}$

$\small \small \small \small ||v||=25$

$\small \small \small ||v||=5$

Correct answer:

$\small \small \small ||v||=5$

Explanation:

To find the length $\small \small ||v||$ of the vector $\small \small \small v=(-\sqrt{5},\sqrt{20})$ , we take the square root of the dot product $\small v\cdot v$ :

$\small \small |v||=\sqrt{v\cdot v}=\sqrt{(-\sqrt{5})^2+\sqrt{20}^2}=\sqrt{5+20}=\sqrt{25}=5$

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Example Question #94 : Dot Product

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 #95 : Dot Product

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 #96 : Dot Product

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 #97 : Dot Product

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

Possible Answers:

<-1,2>

None of the other answers

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

<2,-1>

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

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 #98 : Dot Product

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

Possible Answers:

None of the other answers.

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

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

<1,-1,0>

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

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 #99 : Dot Product

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 #100 : Dot Product

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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Calculus 3 : Dot Product

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #91 : Dot Product

Example Question #92 : Dot Product

Example Question #93 : Dot Product

Example Question #94 : Dot Product

Example Question #95 : Dot Product

Example Question #96 : Dot Product

Example Question #97 : Dot Product

Example Question #98 : Dot Product

Example Question #99 : Dot Product

Example Question #100 : Dot Product

All Calculus 3 Resources

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