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Calculus 3 : Vectors and Vector Operations

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #23 : Distance Between Vectors

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

Possible Answers:

Correct answer:

Explanation:

To find the distance between two vectors $\left \langle x_1,y_1,z_1\right \rangle$ and $\left \langle x_2,y_2,z_2\right \rangle$ , we use the formula $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

Applying to the vectors from the problem statement, we get

$d=\sqrt{(0-0)^2+(3-3)^2+(1-5)^2}=\sqrt{16}=4$

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Example Question #91 : Calculus 3

Find the distance between the vectors $\left \langle 9,3,8\right \rangle$ and $\left \langle 2,-6,1\right \rangle$

Possible Answers:

$\sqrt{111}$

$\sqrt{117}$

$\sqrt{105}$

$\sqrt{107}$

Correct answer:

$\sqrt{107}$

Explanation:

To find the distance between two vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , we use the formula

$d=\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+(a_3-b_3)^2}$

Using the vectors from the problem statement, we get

$d=\sqrt{(9-2)^2+(3-6)^2+(8-1)^2}=\sqrt{107}$

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Example Question #25 : Distance Between Vectors

Find the distance between the vectors $\left \langle 4,5,-9\right \rangle$ and $\left \langle 5,2,-5\right \rangle$

Possible Answers:

$\sqrt{30}$

$\sqrt{31}$

$\sqrt{26}$

Correct answer:

$\sqrt{26}$

Explanation:

To find the distance between two vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , we use the formula

$d=\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+(a_3-b_3)^2}$

Using the vectors from the problem statement, we get

$d=\sqrt{(4-5)^2+(5-2)^2+(-9+5)^2}=\sqrt{26}$

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Example Question #21 : Distance Between Vectors

Find the distance between the vectors $\left \langle 7,0,7\right \rangle$ and $\left \langle 5,7,-5\right \rangle$

Possible Answers:

$\sqrt{173}$

$\sqrt{155}$

$\sqrt{191}$

$\sqrt{197}$

Correct answer:

$\sqrt{197}$

Explanation:

To find the distance between two vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , we use the formula

$d=\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+(a_3-b_3)^2}$

Using the vectors from the problem statement, we get

$d=\sqrt{(7-5)^2+(0-7)^2+(7+5)^2}=\sqrt{197}$

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Example Question #27 : Distance Between Vectors

Find the distance between the vectors $\left \langle 2,-5,4\right \rangle$ and $\left \langle -8,9,3\right \rangle$

Possible Answers:

$\sqrt{155}$

$\sqrt{297}$

$\sqrt{302}$

$\sqrt{299}$

Correct answer:

$\sqrt{297}$

Explanation:

To find the distance between two vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , you use the formula

$d=\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+(a_3-b_3)^2}$

Using the vectors from the problem statement, we get

$d=\sqrt{(2+8)^2+(-5-9)^2+(4-3)^2}=\sqrt{297}$

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Example Question #101 : Calculus 3

Find the distance between the vectors $\left \langle -6,0,6\right \rangle$ and $\left \langle 1,3,5\right \rangle$

Possible Answers:

$\sqrt{59}$

$\sqrt{62}$

$\sqrt{55}$

$\sqrt{57}$

Correct answer:

$\sqrt{59}$

Explanation:

To find the distance between two vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , you use the formula

$d=\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+(a_3-b_3)^2}$

Using the vectors from the problem statement, we get

$d=\sqrt{(-6-1)^2+(0-3)^2+(6-5)^2}=\sqrt{59}$

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Example Question #31 : Distance Between Vectors

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

Possible Answers:

36.52

6.16

28.18

38.00

Correct answer:

36.52

Explanation:

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

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Example Question #102 : Calculus 3

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

Possible Answers:

37.66

4.69

22.00

13.93

Correct answer:

13.93

Explanation:

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

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Example Question #103 : Calculus 3

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

Possible Answers:

21.00

43.60

27.37

4.58

Correct answer:

27.37

Explanation:

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

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Example Question #101 : Calculus 3

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

Possible Answers:

35.69

19.75

46.00

6.78

Correct answer:

35.69

Explanation:

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

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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 #23 : Distance Between Vectors

Example Question #91 : Calculus 3

Example Question #25 : Distance Between Vectors

Example Question #21 : Distance Between Vectors

Example Question #27 : Distance Between Vectors

Example Question #101 : Calculus 3

Example Question #31 : Distance Between Vectors

Example Question #102 : Calculus 3

Example Question #103 : Calculus 3

Example Question #101 : Calculus 3

All Calculus 3 Resources

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