Distance between Vectors

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AP Calculus BC › Distance between Vectors

Questions 1 - 10

Find the distance between the vectors $\left \langle 3,-6,1\right \rangle$ and $\left \langle 2,5,5\right \rangle$

$\sqrt{138}$

$\sqrt{115}$

$\sqrt{120}$

$\sqrt{131}$

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 following formula:

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

Therefore, we get

$d=\sqrt{(3-2)^2+(-6-5)^2+(1-5)^2}=\sqrt{138}$

$\begin{align*}&\text{Find the distance between the two vectors:}\&\overrightarrow{u}=\begin{bmatrix}-15\-3\17\end{bmatrix}\overrightarrow{v}=\begin{bmatrix}12\19\6\end{bmatrix}\end{align*}$

Explanation

$\begin{align*}&\text{The distance between two vectors, }\overrightarrow{u}\text{ and }\overrightarrow{v}\text{ is found}\&\text{by summing the squares of the distances between paired coordinates and}\&\text{taking the square root of this sum:}\&\sqrt{\sum_{i=1}^{n}(\overrightarrow{u}_{i}-\overrightarrow{v}_{i})^2}\&\text{It is similar to finding the diagonal length of a rectangle or rectangular prism.}\&\text{For our vectors }\begin{bmatrix}-15\-3\17\end{bmatrix}\text{ and }\begin{bmatrix}12\19\6\end{bmatrix}\text{ this becomes:}\&\sqrt{[-15-(12)]^2+[-3-(19)]^2+[17-(6)]^2}\&\sqrt{(-27)^2+(-22)^2+(11)^2}\&36.52\end{align*}$

Find the distance between the vectors $\left \langle 3,0,6\right \rangle$ and $\left \langle -4,1,3\right \rangle$

$\sqrt{59}$

$\sqrt{60}$

$\sqrt{58}$

$\sqrt{57}$

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 following formula:

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

Therefore, we get

$d=\sqrt{(3+4)^2+(0-1)^2+(6-3)^2}=\sqrt{59}$

$\begin{align*}&\text{Find the distance between the two vectors:}\&\overrightarrow{u}=\begin{bmatrix}14\18\7\end{bmatrix}\overrightarrow{v}=\begin{bmatrix}11\10\-4\end{bmatrix}\end{align*}$

Explanation

$\begin{align*}&\text{The distance between two vectors, }\overrightarrow{u}\text{ and }\overrightarrow{v}\text{ is found}\&\text{by summing the squares of the distances between paired coordinates and}\&\text{taking the square root of this sum:}\&\sqrt{\sum_{i=1}^{n}(\overrightarrow{u}_{i}-\overrightarrow{v}_{i})^2}\&\text{It is similar to finding the diagonal length of a rectangle or rectangular prism.}\&\text{For our vectors }\begin{bmatrix}14\18\7\end{bmatrix}\text{ and }\begin{bmatrix}11\10\-4\end{bmatrix}\text{ this becomes:}\&\sqrt{[14-(11)]^2+[18-(10)]^2+[7-(-4)]^2}\&\sqrt{(3)^2+(8)^2+(11)^2}\&13.93\end{align*}$

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

$\sqrt{77}$

$\sqrt{80}$

$\sqrt{55}$

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

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

Given the vectors

find the distance between them, .

$\sqrt{6}$

$\sqrt{5}$

$\sqrt{7}$

Explanation

To find the distance between the vectors

we do the following calculation:

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

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

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

$\sqrt{14}$

$\sqrt{15}$

$\sqrt{5}$

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

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

Using the vectors from the problem statement, we get

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

$\begin{align*}&\overrightarrow{u}=\begin{bmatrix}-13\8\-19\end{bmatrix}\overrightarrow{v}=\begin{bmatrix}-9\-19\-17\end{bmatrix}\&\text{Determine the distance between these two vectors.}\end{align*}$

Explanation

$\begin{align*}&\text{The distance between two vectors, }\overrightarrow{u}\text{ and }\overrightarrow{v}\text{ is found}\&\text{by summing the squares of the distances between paired coordinates and}\&\text{taking the square root of this sum:}\&\sqrt{\sum_{i=1}^{n}(\overrightarrow{u}_{i}-\overrightarrow{v}_{i})^2}\&\text{It is similar to finding the diagonal length of a rectangle or rectangular prism.}\&\text{For our vectors }\begin{bmatrix}-13\8\-19\end{bmatrix}\text{ and }\begin{bmatrix}-9\-19\-17\end{bmatrix}\text{ this becomes:}\&\sqrt{[-13-(-9)]^2+[8-(-19)]^2+[-19-(-17)]^2}\&\sqrt{(-4)^2+(27)^2+(-2)^2}\&27.37\end{align*}$

$\begin{align*}&\overrightarrow{u}=\begin{bmatrix}8\-7\18\end{bmatrix}\overrightarrow{v}=\begin{bmatrix}-19\-3\-5\end{bmatrix}\&\text{Determine the distance between these two vectors.}\end{align*}$

Explanation

$\begin{align*}&\text{The distance between two vectors, }\overrightarrow{u}\text{ and }\overrightarrow{v}\text{ is found}\&\text{by summing the squares of the distances between paired coordinates and}\&\text{taking the square root of this sum:}\&\sqrt{\sum_{i=1}^{n}(\overrightarrow{u}_{i}-\overrightarrow{v}_{i})^2}\&\text{It is similar to finding the diagonal length of a rectangle or rectangular prism.}\&\text{For our vectors }\begin{bmatrix}8\-7\18\end{bmatrix}\text{ and }\begin{bmatrix}-19\-3\-5\end{bmatrix}\text{ this becomes:}\&\sqrt{[8-(-19)]^2+[-7-(-3)]^2+[18-(-5)]^2}\&\sqrt{(27)^2+(-4)^2+(23)^2}\&35.69\end{align*}$

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

$\sqrt{94}$

$\sqrt{82}$

$\sqrt{66}$

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

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

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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