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Calculus 3 : Distance between Vectors

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

Example Questions

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:

$\sqrt{62}$

$\sqrt{59}$

$\sqrt{57}$

$\sqrt{55}$

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 #81 : Vectors And Vector Operations

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

Possible Answers:

38.00

36.52

28.18

6.16

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 #82 : Vectors And Vector Operations

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

Possible Answers:

37.66

22.00

4.69

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 #101 : 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

4.58

27.37

43.60

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 #35 : Distance Between Vectors

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

Possible Answers:

35.69

6.78

19.75

46.00

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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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 #31 : Distance Between Vectors

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

Possible Answers:

25.42

24.29

34.00

5.83

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 #32 : Distance Between Vectors

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

Possible Answers:

15.00

24.06

3.87

13.38

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 #91 : Vectors And Vector Operations

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

Possible Answers:

46.84

9.49

8.72

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 #34 : Distance Between Vectors

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

Possible Answers:

33.85

25.34

2.83

8.00

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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Calculus 3 : Distance between Vectors

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #101 : Calculus 3

Example Question #81 : Vectors And Vector Operations

Example Question #82 : Vectors And Vector Operations

Example Question #101 : Calculus 3

Example Question #35 : Distance Between Vectors

Example Question #91 : Vectors And Vector Operations

Example Question #31 : Distance Between Vectors

Example Question #32 : Distance Between Vectors

Example Question #91 : Vectors And Vector Operations

Example Question #34 : Distance Between Vectors

All Calculus 3 Resources

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