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Calculus 3 : Vectors and Vector Operations

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Example Questions

Example Question #91 : Vectors And Vector Operations

$\begin{align*}&\overrightarrow{u}=\begin{bmatrix}12\\-13\\0\end{bmatrix}\overrightarrow{v}=\begin{bmatrix}-2\\6\\9\end{bmatrix}\\&\text{Determine the distance between these two vectors.}\end{align*}$

Possible Answers:

15.17

3.74

14.00

25.26

Correct answer:

25.26

Explanation:

$\begin{align*}&\text{The distance between two vectors, }\overrightarrow{u}\text{ and }\overrightarrow{v}\text{ is found}\\&\text{by summing the squares of the distances between paired coordinates and}\\&\text{taking the square root of this sum:}\\&\sqrt{\sum_{i=1}^{n}(\overrightarrow{u}_{i}-\overrightarrow{v}_{i})^2}\\&\text{It is similar to finding the diagonal length of a rectangle or rectangular prism.}\\&\text{For our vectors }\begin{bmatrix}12\\-13\\0\end{bmatrix}\text{ and }\begin{bmatrix}-2\\6\\9\end{bmatrix}\text{ this becomes:}\\&\sqrt{[12-(-2)]^2+[-13-(6)]^2+[0-(9)]^2}\\&\sqrt{(14)^2+(-19)^2+(-9)^2}\\&25.26\end{align*}$

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Example Question #101 : Calculus 3

$\begin{align*}&\overrightarrow{u}=\begin{bmatrix}-9\\7\\6\end{bmatrix}\overrightarrow{v}=\begin{bmatrix}-14\\-16\\0\end{bmatrix}\\&\text{Determine the distance between these two vectors.}\end{align*}$

Possible Answers:

25.42

34.00

5.83

24.29

Correct answer:

24.29

Explanation:

$\begin{align*}&\text{The distance between two vectors, }\overrightarrow{u}\text{ and }\overrightarrow{v}\text{ is found}\\&\text{by summing the squares of the distances between paired coordinates and}\\&\text{taking the square root of this sum:}\\&\sqrt{\sum_{i=1}^{n}(\overrightarrow{u}_{i}-\overrightarrow{v}_{i})^2}\\&\text{It is similar to finding the diagonal length of a rectangle or rectangular prism.}\\&\text{For our vectors }\begin{bmatrix}-9\\7\\6\end{bmatrix}\text{ and }\begin{bmatrix}-14\\-16\\0\end{bmatrix}\text{ this becomes:}\\&\sqrt{[-9-(-14)]^2+[7-(-16)]^2+[6-(0)]^2}\\&\sqrt{(5)^2+(23)^2+(6)^2}\\&24.29\end{align*}$

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Example Question #102 : Calculus 3

$\begin{align*}&\overrightarrow{u}=\begin{bmatrix}-7\\3\\-11\end{bmatrix}\overrightarrow{v}=\begin{bmatrix}10\\-10\\0\end{bmatrix}\\&\text{Determine the distance between these two vectors.}\end{align*}$

Possible Answers:

13.38

15.00

24.06

3.87

Correct answer:

24.06

Explanation:

$\begin{align*}&\text{The distance between two vectors, }\overrightarrow{u}\text{ and }\overrightarrow{v}\text{ is found}\\&\text{by summing the squares of the distances between paired coordinates and}\\&\text{taking the square root of this sum:}\\&\sqrt{\sum_{i=1}^{n}(\overrightarrow{u}_{i}-\overrightarrow{v}_{i})^2}\\&\text{It is similar to finding the diagonal length of a rectangle or rectangular prism.}\\&\text{For our vectors }\begin{bmatrix}-7\\3\\-11\end{bmatrix}\text{ and }\begin{bmatrix}10\\-10\\0\end{bmatrix}\text{ this becomes:}\\&\sqrt{[-7-(10)]^2+[3-(-10)]^2+[-11-(0)]^2}\\&\sqrt{(-17)^2+(13)^2+(-11)^2}\\&24.06\end{align*}$

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Example Question #103 : Calculus 3

$\begin{align*}&\overrightarrow{u}=\begin{bmatrix}16\\19\\2\end{bmatrix}\overrightarrow{v}=\begin{bmatrix}-15\\-14\\-10\end{bmatrix}\\&\text{Determine the distance between these two vectors.}\end{align*}$

Possible Answers:

8.72

9.49

46.84

76.00

Correct answer:

46.84

Explanation:

$\begin{align*}&\text{The distance between two vectors, }\overrightarrow{u}\text{ and }\overrightarrow{v}\text{ is found}\\&\text{by summing the squares of the distances between paired coordinates and}\\&\text{taking the square root of this sum:}\\&\sqrt{\sum_{i=1}^{n}(\overrightarrow{u}_{i}-\overrightarrow{v}_{i})^2}\\&\text{It is similar to finding the diagonal length of a rectangle or rectangular prism.}\\&\text{For our vectors }\begin{bmatrix}16\\19\\2\end{bmatrix}\text{ and }\begin{bmatrix}-15\\-14\\-10\end{bmatrix}\text{ this becomes:}\\&\sqrt{[16-(-15)]^2+[19-(-14)]^2+[2-(-10)]^2}\\&\sqrt{(31)^2+(33)^2+(12)^2}\\&46.84\end{align*}$

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Example Question #104 : Calculus 3

$\begin{align*}&\text{Find the distance between the two vectors:}\\&\overrightarrow{u}=\begin{bmatrix}-10\\13\\-11\end{bmatrix}\overrightarrow{v}=\begin{bmatrix}18\\-6\\-12\end{bmatrix}\end{align*}$

Possible Answers:

8.00

33.85

2.83

25.34

Correct answer:

33.85

Explanation:

$\begin{align*}&\text{The distance between two vectors, }\overrightarrow{u}\text{ and }\overrightarrow{v}\text{ is found}\\&\text{by summing the squares of the distances between paired coordinates and}\\&\text{taking the square root of this sum:}\\&\sqrt{\sum_{i=1}^{n}(\overrightarrow{u}_{i}-\overrightarrow{v}_{i})^2}\\&\text{It is similar to finding the diagonal length of a rectangle or rectangular prism.}\\&\text{For our vectors }\begin{bmatrix}-10\\13\\-11\end{bmatrix}\text{ and }\begin{bmatrix}18\\-6\\-12\end{bmatrix}\text{ this becomes:}\\&\sqrt{[-10-(18)]^2+[13-(-6)]^2+[-11-(-12)]^2}\\&\sqrt{(-28)^2+(19)^2+(1)^2}\\&33.85\end{align*}$

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Example Question #91 : Vectors And Vector Operations

$\begin{align*}&\text{Find the distance between the two vectors:}\\&\overrightarrow{u}=\begin{bmatrix}8\\1\\-2\end{bmatrix}\overrightarrow{v}=\begin{bmatrix}-3\\10\\-16\end{bmatrix}\end{align*}$

Possible Answers:

21.68

19.95

16.00

4.00

Correct answer:

19.95

Explanation:

$\begin{align*}&\text{The distance between two vectors, }\overrightarrow{u}\text{ and }\overrightarrow{v}\text{ is found}\\&\text{by summing the squares of the distances between paired coordinates and}\\&\text{taking the square root of this sum:}\\&\sqrt{\sum_{i=1}^{n}(\overrightarrow{u}_{i}-\overrightarrow{v}_{i})^2}\\&\text{It is similar to finding the diagonal length of a rectangle or rectangular prism.}\\&\text{For our vectors }\begin{bmatrix}8\\1\\-2\end{bmatrix}\text{ and }\begin{bmatrix}-3\\10\\-16\end{bmatrix}\text{ this becomes:}\\&\sqrt{[8-(-3)]^2+[1-(10)]^2+[-2-(-16)]^2}\\&\sqrt{(11)^2+(-9)^2+(14)^2}\\&19.95\end{align*}$

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Example Question #42 : Distance Between Vectors

$\begin{align*}&\text{Find the distance between the two vectors:}\\&\overrightarrow{u}=\begin{bmatrix}-18\\8\\0\end{bmatrix}\overrightarrow{v}=\begin{bmatrix}-2\\5\\-17\end{bmatrix}\end{align*}$

Possible Answers:

29.29

4.00

23.54

2.00

Correct answer:

23.54

Explanation:

$\begin{align*}&\text{The distance between two vectors, }\overrightarrow{u}\text{ and }\overrightarrow{v}\text{ is found}\\&\text{by summing the squares of the distances between paired coordinates and}\\&\text{taking the square root of this sum:}\\&\sqrt{\sum_{i=1}^{n}(\overrightarrow{u}_{i}-\overrightarrow{v}_{i})^2}\\&\text{It is similar to finding the diagonal length of a rectangle or rectangular prism.}\\&\text{For our vectors }\begin{bmatrix}-18\\8\\0\end{bmatrix}\text{ and }\begin{bmatrix}-2\\5\\-17\end{bmatrix}\text{ this becomes:}\\&\sqrt{[-18-(-2)]^2+[8-(5)]^2+[0-(-17)]^2}\\&\sqrt{(-16)^2+(3)^2+(17)^2}\\&23.54\end{align*}$

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Example Question #43 : Distance Between Vectors

$\begin{align*}&\text{Calculate the distance between the two vectors:}\\&\begin{bmatrix}-1\\-20\\7\end{bmatrix}\text{ and }\begin{bmatrix}-13\\14\\6\end{bmatrix}\end{align*}$

Possible Answers:

36.07

21.00

20.02

4.58

Correct answer:

36.07

Explanation:

$\begin{align*}&\text{The distance between two vectors, }\overrightarrow{u}\text{ and }\overrightarrow{v}\text{ is found}\\&\text{by summing the squares of the distances between paired coordinates and}\\&\text{taking the square root of this sum:}\\&\sqrt{\sum_{i=1}^{n}(\overrightarrow{u}_{i}-\overrightarrow{v}_{i})^2}\\&\text{It is similar to finding the diagonal length of a rectangle or rectangular prism.}\\&\text{For our vectors }\begin{bmatrix}-1\\-20\\7\end{bmatrix}\text{ and }\begin{bmatrix}-13\\14\\6\end{bmatrix}\text{ this becomes:}\\&\sqrt{[-1-(-13)]^2+[-20-(14)]^2+[7-(6)]^2}\\&\sqrt{(12)^2+(-34)^2+(1)^2}\\&36.07\end{align*}$

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Example Question #41 : Distance Between Vectors

Find the distance between the vectors $\left \langle 3,-3,-2\right \rangle$ and $\left \langle 1, 1, 7\right \rangle$

Possible Answers:

$\sqrt{111}$

$\sqrt{105}$

$\sqrt{101}$

Correct answer:

$\sqrt{101}$

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

$d=\sqrt{(3-1)^2+(-3-1)^2+(-2-7)^2}=\sqrt{101}$

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Example Question #111 : Calculus 3

Find the distance between the vectors $\left \langle -5,-1,-3\right \rangle$ and $\left \langle 2, 6, -4\right \rangle$

Possible Answers:

$\sqrt{98}$

$\sqrt{103}$

$\sqrt{99}$

Correct answer:

$\sqrt{99}$

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

$d=\sqrt{(-5-2)^2+(-1-6)^2+(-3+4)^2}=\sqrt{99}$

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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 #91 : Vectors And Vector Operations

Example Question #101 : Calculus 3

Example Question #102 : Calculus 3

Example Question #103 : Calculus 3

Example Question #104 : Calculus 3

Example Question #91 : Vectors And Vector Operations

Example Question #42 : Distance Between Vectors

Example Question #43 : Distance Between Vectors

Example Question #41 : Distance Between Vectors

Example Question #111 : Calculus 3

All Calculus 3 Resources

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