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Calculus 3 : Distance between Vectors

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

Given the vectors

v=(1,0,5)

w=(0,2,4)

find the distance between them, d(v,w) .

Possible Answers:

$\sqrt{7}$

$\sqrt{5}$

$\sqrt{6}$

Correct answer:

$\sqrt{6}$

Explanation:

To find the distance d(v,w) between the vectors

v=(1,0,5)

w=(0,2,4)

we do the following calculation:

$d(v,w)=\sqrt{(1-0)^2+(0-2)^2+(5-4)^2}=\sqrt{1^2+(-2)^2+1^2}$

$=\sqrt{1+4+1}=\sqrt{6}$

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Example Question #2 : Distance Between Vectors

Find the distance d(v,w) between the two vectors

v=(1,3)

w=(5,2)

Possible Answers:

$\sqrt{18}$

$\sqrt{17}$

Correct answer:

$\sqrt{17}$

Explanation:

To find the distance d(v,w) between the two vectors

v=(1,3)

w=(5,2)

we make the following calculation

$d(v,w)=\sqrt{(1-5)^2+(3-2)^2}=\sqrt{16+1}=\sqrt{17}$

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Example Question #1 : Distance Between Vectors

Find the Euclidian distance between the two vectors:

$\mathbf{u}=[5, 3, 0], \, \;\mathbf{v}=[2, -2, \sqrt{2}]$

Possible Answers:

$3\sqrt{2}$

$\frac{9\sqrt{2}}{2}$

Correct answer:

Explanation:

The Euclidian distance between two vectors is:

$d=\sqrt{(u_x-v_x)^2+(u_y-v_y)^2+(u_z-v_z)^2}$

Plugging in the numbers given, we have:

$d=\sqrt{(5-2)^2+(3-(-2))^2+(0-\sqrt2)^2}$

$d=\sqrt{3^2+5^2+\sqrt{2}^2}=\sqrt{9+25+2}=\sqrt{36}=6$

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Example Question #4 : Distance Between Vectors

Find the distance between $\overrightarrow{XY}$ if $X= \left \langle 1,0,-5\right \rangle$ and $Y=\left \langle -3,2,-1\right \rangle$ .

Possible Answers:

$\sqrt{14}$

$3\sqrt2$

$6\sqrt2$

$2\sqrt3$

Correct answer:

Explanation:

Write the formula to find the magnitude of the vector $v=\overrightarrow{XY}$ .

$\left | v\right | = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

Substitute the points into the equation assuming $X=\left \langle x_1,y_1,z_1\right \rangle$ and $Y=\left \langle x_2,y_2,z_2\right \rangle$ .

$\left | v\right | = \sqrt{(-3-1)^2+(2-0)^2+(-1-(-5))^2}$

$\left | v\right | = \sqrt{(-4)^2+(2)^2+4^2}$

$\left | v\right | = \sqrt{16+4+16} = \sqrt{36}= 6$

The answer is:

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Example Question #1 : Distance Between Vectors

Calculate the length of line segment AB given A(−5, −2, 0) and B(6, 0, 3):

Possible Answers:

$\sqrt{134}$

$\sqrt{114}$

$\sqrt{14}$

None of the Above

Correct answer:

$\sqrt{134}$

Explanation:

First we need to find $\vec{AB}$

$\vec{AB}=(6-(-5),0-(-2),3-0)=(11,2,3)$

$\left \| \vec{AB} \right \|= \sqrt{121+4+9}=\sqrt{134}$

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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:

$10\sqrt{2}$

$3\sqrt{14}$

$\sqrt{201}$

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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Calculus 3 : Distance between Vectors

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1 : Distance Between Vectors

Example Question #2 : Distance Between Vectors

Example Question #1 : Distance Between Vectors

Example Question #4 : 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 #1 : Distance Between Vectors

Example Question #1 : Distance Between Vectors

All Calculus 3 Resources

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