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Calculus 3 : Dot Product

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Example Questions

Example Question #81 : Dot Product

Find the dot product: $\left \langle -2,9\right \rangle\cdot \left \langle -3,7\right \rangle$

Possible Answers:

$\left \langle 6,63\right \rangle$

$\left \langle -5,16\right \rangle$

Correct answer:

Explanation:

Write the dot product formula.

$a\cdot b = a_1b_1+a_2b_2$

Substitute the values of the vectors and solve. The dot product will result in a number, not a vector.

(-2)(-3) + (9)(7) = 6 +63 =69

The dot product is:

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Example Question #82 : Dot Product

Find the length $\small ||v||$ of the vector $\small v=(1,\sqrt{3})$ .

Possible Answers:

$\small \small \small ||v||=\sqrt{1+\sqrt{3}}$

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

$\small \small \small ||v||=1+\sqrt{3}$

$\small \small ||v||=2$

Correct answer:

$\small \small ||v||=2$

Explanation:

To find the length $\small ||v||$ of the vector $\small v=(1,\sqrt{3})$ , we take the square root of the dot product $\small v\cdot v$ :

$\small \small ||v||=\sqrt{v\cdot v}=\sqrt{1^2+\sqrt{3}^2}=\sqrt{4}=2$

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Example Question #396 : Vectors And Vector Operations

Find the length $\small ||v||$ of the vector $\small \small v=(15,12)$ .

Possible Answers:

$\small \small \small ||v||=\sqrt{370}$

$\small \small \small ||v||=\sqrt{369}$

$\small \small \small ||v||=27$

$\small \small \small ||v||=369$

Correct answer:

$\small \small \small ||v||=\sqrt{369}$

Explanation:

To find the length $\small ||v||$ of the vector $\small \small v=(15,12)$ , we take the square root of the dot product $\small v\cdot v$ :

$\small \small \small ||v||=\sqrt{v\cdot v}=\sqrt{15^2+12^2}=\sqrt{225+144}=\sqrt{369}$

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Example Question #397 : Vectors And Vector Operations

Find the length $\small ||v||$ of the vector $\small \small \small v=(5,5)$ .

Possible Answers:

$\small \small \small ||v||=10$

$\small \small \small ||v||=2\sqrt{5}$

$\small \small ||v||=5\sqrt{2}$

$\small \small \small ||v||=50$

Correct answer:

$\small \small ||v||=5\sqrt{2}$

Explanation:

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

$\small \small \small \small \small ||v||=\sqrt{v\cdot v}=\sqrt{5^2+5^2}=\sqrt{25+25}=\sqrt{25\cdot 2}=5\sqrt{2}$

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Example Question #398 : Vectors And Vector Operations

Find the length $\small ||v||$ of the vector $\small \small \small \small v=(3,-1)$ .

Possible Answers:

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

$\small \small \small ||v||=2$

$\small \small \small ||v||=10$

$\small \small ||v||=\sqrt{10}$

Correct answer:

$\small \small ||v||=\sqrt{10}$

Explanation:

To find the length $\small ||v||$ of the vector $\small \small \small \small v=(3,-1)$ , we take the square root of the dot product $\small v\cdot v$ :

$\small \small \small \small \small \small ||v||=\sqrt{v\cdot v}=\sqrt{3^2+(-1)^2}=\sqrt{9+1}=\sqrt{10}$

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Example Question #2401 : Calculus 3

Find the length $\small ||v||$ of the vector $\small \small \small \small \small v=(\pi,\pi^4)$ .

Possible Answers:

$\small \small \small ||v||=\sqrt{\pi+\pi^4}$

$\small \small \small ||v||=\pi^2+\pi^6$

$\small \small ||v||=\pi\sqrt{1+\pi^6}$

$\small \small \small ||v||=\pi+\pi^4$

Correct answer:

$\small \small ||v||=\pi\sqrt{1+\pi^6}$

Explanation:

To find the length $\small ||v||$ of the vector $\small \small \small \small \small v=(\pi,\pi^4)$ , we take the square root of the dot product $\small v\cdot v$ :

$\small \small \small \small \small \small \small ||v||=\sqrt{v\cdot v}=\sqrt{\pi^2+(\pi^4)^2}=\sqrt{\pi^2+\pi^8}=\sqrt{\pi^2(1+\pi^6)}=\pi\sqrt{1+\pi^6}$

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Example Question #2402 : Calculus 3

Find the length $\small ||v||$ of the vector $\small \small \small \small \small \small v=(\sqrt{3},\sqrt{5})$ .

Possible Answers:

$\small \small \small ||v||=8$

$\small \small \small ||v||=\sqrt{3}+\sqrt{5}$

$\small \small \small ||v||=\sqrt{\sqrt{3}+\sqrt{5}}$

$\small \small ||v||=2\sqrt{2}$

Correct answer:

$\small \small ||v||=2\sqrt{2}$

Explanation:

To find the length $\small ||v||$ of the vector $\small \small \small \small \small \small v=(\sqrt{3},\sqrt{5})$ , we take the square root of the dot product $\small v\cdot v$ :

$\small \small |v||=\sqrt{v\cdot v}=\sqrt{\sqrt{3}^2+\sqrt{5}^2}=\sqrt{3+5}=\sqrt{8}=\sqrt{4\cdot 2}=2\sqrt{2}$

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Example Question #2403 : Calculus 3

Find the length $\small ||v||$ of the vector $\small \small \small \small \small \small \small v=(1,3,7)$ .

Possible Answers:

$\small \small ||v||=\sqrt{59}$

$\small \small \small ||v||=\sqrt{11}$

$\small \small \small ||v||=11$

$\small \small \small ||v||=59$

Correct answer:

$\small \small ||v||=\sqrt{59}$

Explanation:

To find the length $\small ||v||$ of the vector $\small \small \small \small \small \small \small v=(1,3,7)$ , we take the square root of the dot product $\small v\cdot v$ :

$\small \small \small |v||=\sqrt{v\cdot v}=\sqrt{1^2+3^2+7^2}=\sqrt{1+9+49}=\sqrt{59}$

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Example Question #2404 : Calculus 3

Find the length $\small ||v||$ of the vector $\small \small \small \small \small \small \small \small v=(\pi,\pi^2,\pi^6)$ .

Possible Answers:

$\small \small \small ||v||=\sqrt{\pi^2+\pi^{12}}$

$\small \small \small ||v||=\pi^2+\pi^4+\pi^{12}$

$\small \small \small ||v||=1+\pi^2+\pi^{10}$

$\small \small ||v||=\pi\sqrt{1+\pi^2+\pi^{10}}$

Correct answer:

$\small \small ||v||=\pi\sqrt{1+\pi^2+\pi^{10}}$

Explanation:

To find the length $\small ||v||$ of the vector $\small \small \small \small \small \small \small \small v=(\pi,\pi^2,\pi^6)$ , we take the square root of the dot product $\small v\cdot v$ :

$\small \small \small \small \small |v||=\sqrt{v\cdot v}=\sqrt{\pi^2+(\pi^2)^2+(\pi^6])^2}=\sqrt{\pi^2+\pi^4+\pi^{12}}=\sqrt{\pi^2(1+\pi^2+\pi^{10})}$

$\small =\pi\sqrt{1+\pi^2+\pi^{10}}$

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Example Question #2405 : Calculus 3

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

Possible Answers:

$\small \small \small \small ||v||=\frac{1}{2}(e+\pi+1)$

$\small \small \small \small ||v||=e+\pi+1$

$\small \small \small \small ||v||=e^2+\pi^2+1$

$\small \small \small ||v||=\sqrt{e^2+\pi^2+1}$

Correct answer:

$\small \small \small ||v||=\sqrt{e^2+\pi^2+1}$

Explanation:

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

$\small \small \small \small \small |v||=\sqrt{v\cdot v}=\sqrt{e^2+\pi^2+1^2}=\sqrt{e^2+\pi^2+1}$

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Calculus 3 : Dot Product

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #81 : Dot Product

Example Question #82 : Dot Product

Example Question #396 : Vectors And Vector Operations

Example Question #397 : Vectors And Vector Operations

Example Question #398 : Vectors And Vector Operations

Example Question #2401 : Calculus 3

Example Question #2402 : Calculus 3

Example Question #2403 : Calculus 3

Example Question #2404 : Calculus 3

Example Question #2405 : Calculus 3

All Calculus 3 Resources

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