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Calculus 2 : Vector Form

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #84 : Vectors & Spaces

What is the vector form of ?

Possible Answers:

$\left \langle 0,0,1\right \rangle$

$\left \langle -1,0,0\right \rangle$

$\left \langle 0,0,-1\right \rangle$

$\left \langle 1,0,0\right \rangle$

$\left \langle 0,-1,0\right \rangle$

Correct answer:

$\left \langle 0,0,-1\right \rangle$

Explanation:

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ .

So for , we can derive the vector form $\left \langle 0,0,-1\right \rangle$ .

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Example Question #85 : Vectors & Spaces

Given points (2,0,-10) and (-5,1,6) , what is the vector form of the distance between the points?

Possible Answers:

$\left \langle -7,1,16\right \rangle$

$\left \langle 7,-1,16\right \rangle$

$\left \langle -7,-1,16\right \rangle$

$\left \langle 7,1,-16\right \rangle$

$\left \langle 7,1,16\right \rangle$

Correct answer:

$\left \langle -7,1,16\right \rangle$

Explanation:

In order to derive the vector form of the distance between two points, we must find the difference between the , , and elements of the points. That is, for any point $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , the distance is the vector $v=\left \langle b_1-a_1,b_2-a_2,b_3-a_3\right \rangle$ .

Subbing in our original points (2,0,-10) and (-5,1,6) , we get:

$v=\left \langle -5-2,1-0,6-(-10)\right \rangle$

$v=\left \langle -7,1,16\right \rangle$

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Example Question #81 : Vectors & Spaces

Given points (5,-5,5) and (-4,4,-4) , what is the vector form of the distance between the points?

Possible Answers:

$\left \langle -9,-9,-9\right \rangle$

$\left \langle 9,-9,-9\right \rangle$

$\left \langle -9,-9,9\right \rangle$

$\left \langle -9,9,-9\right \rangle$

$\left \langle 9,9,9\right \rangle$

Correct answer:

$\left \langle -9,9,-9\right \rangle$

Explanation:

Subbing in our original points (5,-5,5) and (-4,4,-4) , we get:

$v=\left \langle -4-5,4-(-5),-4-5\right \rangle$

$v=\left \langle -9,9,-9\right \rangle$

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Example Question #87 : Vectors & Spaces

What is the vector form of 8i-9j+10k ?

Possible Answers:

$\left \langle 8,-9,10\right \rangle$

$\left \langle -8,9,10\right \rangle$

$\left \langle 8,9,10\right \rangle$

$\left \langle 8,9,-10\right \rangle$

$\left \langle -8,-9,10\right \rangle$

Correct answer:

$\left \langle 8,-9,10\right \rangle$

Explanation:

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients. That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ . So for 8i-9j+10k , we can derive the vector form $\left \langle 8,-9,10\right \rangle$ .

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Example Question #431 : Parametric, Polar, And Vector

What is the vector form of i+3j+k ?

Possible Answers:

$\left \langle -1,-3,-1\right \rangle$

$\left \langle -1,-\frac{1}{3},-1\right \rangle$

$\left \langle 1,-3,1\right \rangle$

$\left \langle 1,\frac{1}{3},1\right \rangle$

$\left \langle 1,3,1\right \rangle$

Correct answer:

$\left \langle 1,3,1\right \rangle$

Explanation:

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients. That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ . So for i+3j+k , we can derive the vector form $\left \langle 1,3,1\right \rangle$ .

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Example Question #81 : Vector Form

Given points (5,4,3) and (-3,-4,-5) , what is the vector form of the distance between the points?

Possible Answers:

$\left \langle -8,-8,-8\right \rangle$

$\left \langle 8,8,8\right \rangle$

$\left \langle -8,-8,8\right \rangle$

$\left \langle 8,-8,8\right \rangle$

$\left \langle 8,-8,-8\right \rangle$

Correct answer:

$\left \langle -8,-8,-8\right \rangle$

Explanation:

Subbing in our original points (5,4,3) and (-3,-4,-5) , we get:

$v=\left \langle -3-5,-4-4,-5-3\right \rangle$

$v=\left \langle -8,-8,-8\right \rangle$

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Example Question #101 : Linear Algebra

Given points (2,7,-2) and (4,14,-7) , what is the vector form of the distance between the points?

Possible Answers:

$\left \langle 7,2,-9\right \rangle$

$\left \langle 2,7,-5\right \rangle$

$\left \langle 2,7,9\right \rangle$

$\left \langle 2,7,5\right \rangle$

$\left \langle 2,7,-9\right \rangle$

Correct answer:

$\left \langle 2,7,-5\right \rangle$

Explanation:

Subbing in our original points (2,7,-2) and (4,14,-7) , we get:

$v=\left \langle 4-2,14-7,-7-(-2)\right \rangle$

$v=\left \langle 2,7,-5\right \rangle$

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Example Question #434 : Parametric, Polar, And Vector

What is the vector form of 10i-j+10k ?

Possible Answers:

$\left \langle 10,1,10\right \rangle$

$\left \langle -1,10,10\right \rangle$

$\left \langle 10,10,1\right \rangle$

$\left \langle 10,-1,10\right \rangle$

$\left \langle 10,10,-1\right \rangle$

Correct answer:

$\left \langle 10,-1,10\right \rangle$

Explanation:

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ .

So for 10i-j+10k , we can derive the vector form $\left \langle 10,-1,10\right \rangle$ .

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Example Question #84 : Vector

What is the vector form of i+j ?

Possible Answers:

$\left \langle 1,-1,0\right \rangle$

$\left \langle 1,1,0\right \rangle$

$\left \langle 0,1,1\right \rangle$

$\left \langle -1,1,0\right \rangle$

$\left \langle 1,0,1\right \rangle$

Correct answer:

$\left \langle 1,1,0\right \rangle$

Explanation:

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ .

So for i+j , we can derive the vector form $\left \langle 1,1,0\right \rangle$ .

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Example Question #91 : Vectors

Given points (0,8,-1) and (2,2,2) , what is the vector form of the distance between the points?

Possible Answers:

$\left \langle 2,6,3\right \rangle$

$\left \langle 2,-6,-3\right \rangle$

$\left \langle -2,6,3\right \rangle$

$\left \langle 2,-6,3\right \rangle$

$\left \langle 2,6,-3\right \rangle$

Correct answer:

$\left \langle 2,-6,3\right \rangle$

Explanation:

In order to derive the vector form of the distance between two points, we must find the difference between the , , and elements of the points.

That is, for any point $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , the distance is the vector $v=\left \langle b_1-a_1,b_2-a_2,b_3-a_3\right \rangle$ .

Subbing in our original points (0,8,-1) and (2,2,2) , we get:

$v=\left \langle 2-0,2-8,2-(-1)\right \rangle$

$v=\left \langle 2,-6,3\right \rangle$

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Calculus 2 : Vector Form

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #84 : Vectors & Spaces

Example Question #85 : Vectors & Spaces

Example Question #81 : Vectors & Spaces

Example Question #87 : Vectors & Spaces

Example Question #431 : Parametric, Polar, And Vector

Example Question #81 : Vector Form

Example Question #101 : Linear Algebra

Example Question #434 : Parametric, Polar, And Vector

Example Question #84 : Vector

Example Question #91 : Vectors

All Calculus 2 Resources

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