Distance between Vectors - AP Calculus BC

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Question

Given the vectors

find the distance between them, .

Answer

To find the distance between the vectors

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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Question

Find the distance between the two vectors

Answer

To find the distance between the two vectors

we make the following calculation

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

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Question

Find the Euclidian distance between the two vectors:

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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

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

so

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

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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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 $\sqrt{(4-4)^2+(11-1)^2+(10-0)^2}=\sqrt{200}$

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Question

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

Answer

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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Question

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

Answer

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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Question

Find the distance between the vectors $\left \langle 13,7,12\right \rangle$ and $\left \langle 8,7,10\right \rangle$

Answer

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

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

Using the vectors from the problem statement, we get

$d=\sqrt{(13-8)^2+(7-7)^2+(12-10)^2}=\sqrt{29}$

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Question

Find the distance between the vectors $\left \langle 10,5,6\right \rangle$ and $\left \langle 9,8,7\right \rangle$

Answer

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

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

Using the vectors from the problem statement, we get

$d=\sqrt{(10-9)^2+(5-8)^2+(6-7)^2}=\sqrt{11}$

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Question

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

Answer

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

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

Using the vectors from the problem statement, we get

$d=\sqrt{(3-1)^2+(4-5)^2+(5-2)^2}=\sqrt{14}$

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Question

Find the distance between the vectors $\left \langle 0,7,11\right \rangle$ and $\left \langle -3,-2,9\right \rangle$

Answer

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

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

Using the vectors from the problem statement, we get

$d=\sqrt{(0+3)^2+(7+2)^2+(11-9)^2}=\sqrt{94}$

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Question

Find the distance between the two vectors u and v:

$\mathbf{u}=13\mathbf{i}+8\mathbf{j}+2\sqrt{2}\mathbf{k}$

$\mathbf{v}=9\mathbf{i}+3\mathbf{j}$

Answer

The Euclidean distance between two vectors:

$\mathbf{u}=u_x\mathbf{i}+u_y\mathbf{j}+u_z\mathbf{k}$ and $\mathbf{v}=v_x\mathbf{i}+v_y\mathbf{j}+v_z\mathbf{k}$

is given by:

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

For the given vectors, this results in the distance:

$d=\sqrt{(13-9)^2+(8-3)^2+(2\sqrt{2}-0)^2}$

$d=\sqrt{4^2+5^2+(2\sqrt{2})^2}=\sqrt{16+25+8}=\sqrt{49}=7$

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