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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 #71 : Vectors And Vector Operations

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

Possible Answers:

$\sqrt{20}$ $\displaystyle \sqrt{20}$

$\sqrt{29}$ $\displaystyle \sqrt{29}$

$\displaystyle 6$

$\displaystyle 5$

Correct answer:

$\sqrt{29}$ $\displaystyle \sqrt{29}$

Explanation:

To find the distance between two vectors $a=\left \langle x_1,y_1,z_1\right \rangle$ $\displaystyle a=\left \langle x_1,y_1,z_1\right \rangle$ and $b=\left \langle x_2,y_2,z_2\right \rangle$ $\displaystyle b=\left \langle x_2,y_2,z_2\right \rangle$, we use the formula

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

Using the vectors from the problem statement, we get

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

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Example Question #72 : Vectors And Vector Operations

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

Possible Answers:

$\displaystyle 7$

$\sqrt{12}$ $\displaystyle \sqrt{12}$

$\sqrt{11}$ $\displaystyle \sqrt{11}$

$\displaystyle 5$

Correct answer:

$\sqrt{11}$ $\displaystyle \sqrt{11}$

Explanation:

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

Using the vectors from the problem statement, we get

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

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Example Question #13 : Distance Between Vectors

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

Possible Answers:

$\displaystyle 4$

$\sqrt{14}$ $\displaystyle \sqrt{14}$

$\sqrt{15}$ $\displaystyle \sqrt{15}$

$\sqrt{5}$ $\displaystyle \sqrt{5}$

Correct answer:

$\sqrt{14}$ $\displaystyle \sqrt{14}$

Explanation:

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

Using the vectors from the problem statement, we get

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

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Example Question #19 : Distance Between Vectors

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

Possible Answers:

$\sqrt{66}$ $\displaystyle \sqrt{66}$

$\displaystyle 9$

$\sqrt{82}$ $\displaystyle \sqrt{82}$

$\sqrt{94}$ $\displaystyle \sqrt{94}$

Correct answer:

$\sqrt{94}$ $\displaystyle \sqrt{94}$

Explanation:

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

Using the vectors from the problem statement, we get

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

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Example Question #71 : Vectors And Vector Operations

Find the distance between the two vectors u and v:

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

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

Possible Answers:

$\displaystyle 12$

$\displaystyle 15$

$\displaystyle 23$

$\displaystyle 7$

$\displaystyle 4$

Correct answer:

$\displaystyle 7$

Explanation:

The Euclidean distance between two vectors:

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

is given by:

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

For the given vectors, this results in the distance:

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

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

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Example Question #21 : Distance Between Vectors

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

Possible Answers:

$\sqrt{115}$ $\displaystyle \sqrt{115}$

$\sqrt{131}$ $\displaystyle \sqrt{131}$

$\sqrt{120}$ $\displaystyle \sqrt{120}$

$\sqrt{138}$ $\displaystyle \sqrt{138}$

Correct answer:

$\sqrt{138}$ $\displaystyle \sqrt{138}$

Explanation:

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

Therefore, we get

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

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Example Question #22 : Distance Between Vectors

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

Possible Answers:

$\sqrt{58}$ $\displaystyle \sqrt{58}$

$\sqrt{57}$ $\displaystyle \sqrt{57}$

$\sqrt{59}$ $\displaystyle \sqrt{59}$

$\sqrt{60}$ $\displaystyle \sqrt{60}$

Correct answer:

$\sqrt{59}$ $\displaystyle \sqrt{59}$

Explanation:

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

Therefore, we get

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

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Example Question #21 : Distance Between Vectors

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

Possible Answers:

$\sqrt{11}$ $\displaystyle \sqrt{11}$

$\sqrt{10}$ $\displaystyle \sqrt{10}$

$\sqrt{13}$ $\displaystyle \sqrt{13}$

$\displaystyle 3$

Correct answer:

$\sqrt{11}$ $\displaystyle \sqrt{11}$

Explanation:

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

Using the vectors form the problem statement and applying, we get

$d=\sqrt{(3-2)^2+(0-1)^2+(4-1)^2}=\sqrt{11}$ $\displaystyle d=\sqrt{(3-2)^2+(0-1)^2+(4-1)^2}=\sqrt{11}$

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Example Question #24 : Distance Between Vectors

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

Possible Answers:

$\sqrt{30}$ $\displaystyle \sqrt{30}$

$\sqrt{19}$ $\displaystyle \sqrt{19}$

$\sqrt{12}$ $\displaystyle \sqrt{12}$

$\sqrt{18}$ $\displaystyle \sqrt{18}$

Correct answer:

$\sqrt{30}$ $\displaystyle \sqrt{30}$

Explanation:

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

Using the vectors form the problem statement and applying, we get

$d=\sqrt{(5-3)^2+(5-0)^2+(2-1)^2}=\sqrt{30}$ $\displaystyle d=\sqrt{(5-3)^2+(5-0)^2+(2-1)^2}=\sqrt{30}$

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Example Question #25 : Distance Between Vectors

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

Possible Answers:

$\sqrt{45}$ $\displaystyle \sqrt{45}$

$\sqrt{41}$ $\displaystyle \sqrt{41}$

$\sqrt{42}$ $\displaystyle \sqrt{42}$

$\sqrt{37}$ $\displaystyle \sqrt{37}$

Correct answer:

$\sqrt{42}$ $\displaystyle \sqrt{42}$

Explanation:

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

Applying to the vectors from the problem statement, we get

$d=\sqrt{(2-3)^2+(1-6)^2+(1-5)^2}=\sqrt{42}$ $\displaystyle d=\sqrt{(2-3)^2+(1-6)^2+(1-5)^2}=\sqrt{42}$

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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 #71 : Vectors And Vector Operations

Example Question #72 : Vectors And Vector Operations

Example Question #13 : Distance Between Vectors

Example Question #19 : Distance Between Vectors

Example Question #71 : Vectors And Vector Operations

Example Question #21 : Distance Between Vectors

Example Question #22 : Distance Between Vectors

Example Question #21 : Distance Between Vectors

Example Question #24 : Distance Between Vectors

Example Question #25 : Distance Between Vectors

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