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Calculus 3 : Vectors and Vector Operations

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 #1 : Distance Between Vectors

Compute the distance between the vectors $\left \langle 2,3,5\right \rangle$ and $\left \langle 2,0,9\right \rangle$ .

Possible Answers:

$\sqrt8$

Correct answer:

Explanation:

The formula for the distance between two vectors

$a=\left \langle x_1,y_1,z_1\right \rangle$

and

$b=\left \langle x_2,y_2,z_2\right \rangle$ is

$distance=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$ .

Using the given vectors, we get

$\sqrt{(2-2)^2+(3-0)^2+(5-9)^2}=\sqrt{0+9+16}=5$

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Example Question #1 : Distance Between Vectors

Find the distance between the vectors $\left \langle 0,1,4\right \rangle$ and $\left \langle 2,9,1\right \rangle$ .

Possible Answers:

$\sqrt{77}$

$\sqrt{80}$

$\sqrt{55}$

Correct answer:

$\sqrt{77}$

Explanation:

To find the distance between the vectors, we use the formula

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$ ,

where one vector is

$V_1=\left \langle x_1,y_1,z_1\right \rangle$

and the other is

$V_2=\left \langle x_2,y_2,z_2\right \rangle$ .

Using the vectors we were given, we get

$d=\sqrt{(0-2)^2+(1-9)^2+(4-1)^2}=\sqrt{4+64+9}=\sqrt{77}$ .

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Example Question #1 : Distance Between Vectors

Find the distance between the vectors $\left \langle 1,3,3\right \rangle$ and $\left \langle 2,-1,1\right \rangle$ .

Possible Answers:

$\sqrt{21}$

$\sqrt{22}$

$\sqrt{19}$

Correct answer:

$\sqrt{21}$

Explanation:

To find the distance between the vectors, we use the formula

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$ ,

where one vector is

$V_1=\left \langle x_1,y_1,z_1\right \rangle$

and the other is

$V_2=\left \langle x_2,y_2,z_2\right \rangle$ .

Using the vectors we were given, we get

$d=\sqrt{(1-2)^2+(3-(-1))^2+(3-1)^2}=\sqrt{1+16+4}=\sqrt{21}$ .

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Example Question #1 : Distance Between Vectors

The distance between (1, 3, -10) and (2, 5, 4) is

Possible Answers:

$\sqrt{201}$

$3\sqrt{14}$

$10\sqrt{2}$

None of the Above

Correct answer:

$\sqrt{201}$

Explanation:

We use the distance formula:

$d=\sqrt{(2-1)^2+(5-3)^2+(4+10)^2}=\sqrt{1+4+196}=\sqrt{201}$

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Example Question #1 : Distance Between Vectors

Given the points P:(2,−1,5) andQ:(−2,0,3). The magnitude of the vector from P to Q is:

Possible Answers:

$\left \|\vec{PQ} \right \|=\sqrt{5}$

$\left \|\vec{PQ} \right \|=9$

$\left \|\vec{PQ} \right \|=\sqrt{21}$

None of the Above

$\left \|\vec{PQ} \right \|=\sqrt{65}$

Correct answer:

$\left \|\vec{PQ} \right \|=\sqrt{21}$

Explanation:

First we need to find $\vec{PQ}$ :

$\vec{PQ}=(-2-2,0+1,3-5)=(-4,1,-2)$

Therefore,

$\left \| \vec{PQ} \right \|=\sqrt{16+1+4}=\sqrt{21}$

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Example Question #11 : Distance Between Vectors

Find the distance between the vectors $\left \langle 2,2,4\right \rangle$ and $\left \langle 3,0,9\right \rangle$ .

Possible Answers:

$\sqrt{30}$

Correct answer:

$\sqrt{30}$

Explanation:

To find the distance between two vectors $a=\left \langle x_1,y_1,z_1\right \rangle$ and $b=\left \langle x_2,y_2,z_2\right \rangle$ is $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$ . Using the vectors given, we get $\sqrt{(2-3)^2+(2-0)^2+(4-9)^2}=\sqrt{1+4+25}=\sqrt{30}$

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Example Question #11 : Distance Between Vectors

Find the distance between the vectors $\left \langle 5,3,8\right \rangle$ and $\left \langle 3,0,9\right \rangle$

Possible Answers:

$\sqrt{11}$

$\sqrt{13}$

$\sqrt{14}$

Correct answer:

$\sqrt{14}$

Explanation:

The formula for the distance between vectors $a=\left \langle a,b,c\right \rangle$ and $b=\left \langle x,y,z\right \rangle$ is $d=\sqrt{(a-x)^2+(b-y)^2+(c-z)^2}$ . Using the vectors from the problem statement, we get $\sqrt{(5-3)^2+(3-0)^2+(8-9)^2}=\sqrt{14}$

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Example Question #12 : Distance Between Vectors

Find the distance between the vectors $\left \langle 4,11,10\right \rangle$ and $\left \langle 4,1,0\right \rangle$

Possible Answers:

$\sqrt{200}$

Correct answer:

$\sqrt{200}$

Explanation:

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Example Question #14 : Distance Between Vectors

Find the distance between the vectors $\left \langle 2,2,4\right \rangle$ and $\left \langle 1,-4,1\right \rangle$

Possible Answers:

$\sqrt{10}$

$\sqrt{46}$

Correct answer:

$\sqrt{46}$

Explanation:

To find the distance between two vectors $a=\left \langle x_1,y_1,z_1\right \rangle$ and $b=\left \langle x_2,y_2,z_2\right \rangle$ , we use the formula $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$ . Plugging in the vectors from the problem statement, we get $d=\sqrt{(2-1)^2+(2+4)^2+(4-1)^2}=\sqrt{1+36+9}=\sqrt{46}$

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Example Question #15 : Distance Between Vectors

Find the distance between the vectors $\left \langle -3,1,-4\right \rangle$ and $\left \langle 2,-5,0\right \rangle$

Possible Answers:

$\sqrt{66}$

$\sqrt{77}$

Correct answer:

$\sqrt{77}$

Explanation:

o find the distance between two vectors $a=\left \langle x_1,y_1,z_1\right \rangle$ and $b=\left \langle x_2,y_2,z_2\right \rangle$ , we use the formula $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$ . Plugging in the vectors from the problem statement, we get $d=\sqrt{(-3-2)^2+(1+5)^2+(-4-0)^2}=\sqrt{25+36+16}=\sqrt{77}$

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Calculus 3 : Vectors and Vector Operations

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1 : Distance Between Vectors

Example Question #1 : Distance Between Vectors

Example Question #1 : Distance Between Vectors

Example Question #1 : Distance Between Vectors

Example Question #1 : Distance Between Vectors

Example Question #11 : Distance Between Vectors

Example Question #11 : Distance Between Vectors

Example Question #12 : Distance Between Vectors

Example Question #14 : Distance Between Vectors

Example Question #15 : Distance Between Vectors

All Calculus 3 Resources

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