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Calculus 2 : Vector Calculations

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Example Question #51 : Vector Calculations

What is the cross product of $a=\left \langle -1,0,8\right \rangle$ and $b=\left \langle -8,-1, 0 \right \rangle$ ?

Possible Answers:

$\left \langle -8,-64, 1 \right \rangle$

$\left \langle -8,64, 1 \right \rangle$

$\left \langle 8,64, 1 \right \rangle$

$\left \langle -8,64, -1 \right \rangle$

$\left \langle -8,-64, -1 \right \rangle$

Correct answer:

$\left \langle 8,64, 1 \right \rangle$

Explanation:

In order to find the cross product of two three-dimensional vectors, we must find the determinant of a matrix comprised of the vectors' elements. That is, if $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ and $b=\left \langle b_{1},b_{2},b_{3}\right \rangle$ , then

$a\times b=\begin{Bmatrix} i&j &k \\ a_{1}&a_{2} &a_{3} \\ b_{1}&b_{2} &b_{3} \end{Bmatrix}=\left \langle a_{2}b_{3}-a_{3}b_{2}, a_{1}b_{3}-a_{3}b_{1},a_{1}b_{2}-a_{2}b_{1} \right \rangle$ .

Given $a=\left \langle -1,0,8\right \rangle$ and $b=\left \langle -8,-1, 0 \right \rangle$ , the cross product $a \times b$ is:

$\begin{Bmatrix} i&j &k \\ -1&0 &8 \\ -8&-1 &0 \end{Bmatrix}$

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

$=\left \langle 0+8, 0+64, 1-0 \right \rangle$

$=\left \langle 8,64, 1 \right \rangle$

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Example Question #1062 : Calculus Ii

What is the norm of $a=\left \langle 2,4,6\right \rangle$ ?

Possible Answers:

$\sqrt{55}$

$3\sqrt{6}$

$3\sqrt{7}$

$2\sqrt{15}$

$2\sqrt{14}$

Correct answer:

$2\sqrt{14}$

Explanation:

In order to find the norm of a vector, we must first find the sum of the squares of the vector's elements and take the square root of that sum. Given $a=\left \langle 2,4,6\right \rangle$ , then:

$\left \| a\right \|=\sqrt{(2)^{2}+(4)^{2}+(6)^{2}}$

$\left \| a\right \|=\sqrt{4+16+36}$

$\left \| a\right \|=\sqrt{20+36}$

$\left \| a\right \|=\sqrt{56}$

$\left \| a\right \|=\sqrt{4\times 14}$

$\left \| a\right \|=2\sqrt{14}$

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Example Question #53 : Vector Calculations

What is the norm of $a=\left \langle 3,5,7\right \rangle$ ?

Possible Answers:

$\sqrt{83}$

$3\sqrt{10}$

$3\sqrt{11}$

Correct answer:

$\sqrt{83}$

Explanation:

In order to find the norm of a vector, we must first find the sum of the squares of the vector's elements and take the square root of that sum. Given $a=\left \langle 3,5,7\right \rangle$ , then:

$\left \| a\right \|=\sqrt{(3)^{2}+(5)^{2}+(7)^{2}}$

$\left \| a\right \|=\sqrt{9+25+49}$

$\left \| a\right \|=\sqrt{34+49}$

$\left \| a\right \|=\sqrt{83}$

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Example Question #54 : Vector Calculations

What is the norm of $a=\left \langle9,3,6\right \rangle$ ?

Possible Answers:

$3\sqrt{13}$

$3\sqrt{14}$

$3\sqrt{11}$

$3\sqrt{10}$

Correct answer:

$3\sqrt{14}$

Explanation:

In order to find the norm of a vector, we must first find the sum of the squares of the vector's elements and take the square root of that sum. Given $a=\left \langle9,3,6\right \rangle$ , then:

$\left \| a\right \|=\sqrt{(9)^{2}+(3)^{2}+(6)^{2}}$

$\left \| a\right \|=\sqrt{81+9+36}$

$\left \| a\right \|=\sqrt{90+36}$

$\left \| a\right \|=\sqrt{126}$

$\left \| a\right \|=\sqrt{9\times 14}$

$\left \| a\right \|=3\sqrt{14}$

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Example Question #55 : Vector Calculations

Find the cross product of $a=\left \langle 5,7,2\right \rangle$ and $b=\left \langle 4,4,8\right \rangle$ .

Possible Answers:

$\left \langle 48, 32, -8\right \rangle$

$\left \langle 20, -32, -8\right \rangle$

$\left \langle 20, 32, 8\right \rangle$

$\left \langle -20, -32, -8\right \rangle$

None of the above

Correct answer:

$\left \langle 48, 32, -8\right \rangle$

Explanation:

$a\times b=\begin{Bmatrix} i&j &k \\ a_{1}&a_{2} &a_{3} \\ b_{1}&b_{2} &b_{3} \end{Bmatrix}=\left \langle a_{2}b_{3}-a_{3}b_{2}, a_{1}b_{3}-a_{3}b_{1},a_{1}b_{2}-a_{2}b_{1} \right \rangle$ .

Given $a=\left \langle 5,7,2\right \rangle$ and $b=\left \langle 4,4,8\right \rangle$ . the cross product $a \times b$ is:

$\begin{Bmatrix} i&j &k \\ 5&7 &2 \\ 4&4 &8 \end{Bmatrix}$

$=\left \langle (7)(8)-(2)(4), (5)(8)-(2)(4), (5)(4)-(7)(4) \right \rangle$

$=\left \langle 56-8, 40-8, 20-28\right \rangle$

$=\left \langle 48, 32, -8\right \rangle$

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Example Question #56 : Vector Calculations

What is the cross product of $a=\left \langle 9,8,7\right \rangle$ and $b=\left \langle 0,1,2\right \rangle$ ?

Possible Answers:

$\left \langle -9, 18, 9 \right \rangle$

$\left \langle -9, -18, 9 \right \rangle$

$\left \langle -9, -18, -9 \right \rangle$

$\left \langle 9, 18, 9 \right \rangle$

$\left \langle -9, 18, -9 \right \rangle$

Correct answer:

$\left \langle 9, 18, 9 \right \rangle$

Explanation:

$a\times b=\begin{Bmatrix} i&j &k \\ a_{1}&a_{2} &a_{3} \\ b_{1}&b_{2} &b_{3} \end{Bmatrix}=\left \langle a_{2}b_{3}-a_{3}b_{2}, a_{1}b_{3}-a_{3}b_{1},a_{1}b_{2}-a_{2}b_{1} \right \rangle$ .

Given $a=\left \langle 9,8,7\right \rangle$ and $b=\left \langle 0,1,2\right \rangle$ , the cross product $a \times b$ is:

$\begin{Bmatrix} i&j &k \\ 9&8 &7 \\ 0&1 &2 \end{Bmatrix}$

$=\left \langle (8)(2)-(7)(1), (9)(2)-(7)(0), (9)(1)-(8)(0) \right \rangle$

$=\left \langle 16-7, 18-0, 9-0 \right \rangle$

$=\left \langle 9, 18, 9 \right \rangle$

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Example Question #57 : Vector Calculations

What is the cross product of $a=\left \langle 7,-4,3\right \rangle$ and $b=\left \langle 5,-5,5\right \rangle$ ?

Possible Answers:

$\left \langle -5, 20, -15 \right \rangle$

$\left \langle -5,-20, -15 \right \rangle$

$\left \langle 5, 20, 15 \right \rangle$

$\left \langle 5, 20, -15 \right \rangle$

$\left \langle -5, 20, 15 \right \rangle$

Correct answer:

$\left \langle -5, 20, -15 \right \rangle$

Explanation:

$a\times b=\begin{Bmatrix} i&j &k \\ a_{1}&a_{2} &a_{3} \\ b_{1}&b_{2} &b_{3} \end{Bmatrix}=\left \langle a_{2}b_{3}-a_{3}b_{2}, a_{1}b_{3}-a_{3}b_{1},a_{1}b_{2}-a_{2}b_{1} \right \rangle$ .

Given $a=\left \langle 7,-4,3\right \rangle$ and $b=\left \langle 5,-5,5\right \rangle$ , the cross product $a \times b$ is:

$\begin{Bmatrix} i&j &k \\ 7&-4 &3 \\ 5&-5 &5 \end{Bmatrix}$

$=\left \langle (-4)(5)-(3)(-5), (7)(5)-(3)(5), (7)(-5)-(-4)(5) \right \rangle$

$=\left \langle -20+15, 35-15, -35+20 \right \rangle$

$=\left \langle -5, 20, -15 \right \rangle$

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Example Question #58 : Vector Calculations

What is the cross product of $a=\left \langle 0,-3,3\right \rangle$ and $b=\left \langle -3,0,3\right \rangle$ ?

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:

$a\times b=\begin{Bmatrix} i&j &k \\ a_{1}&a_{2} &a_{3} \\ b_{1}&b_{2} &b_{3} \end{Bmatrix}=\left \langle a_{2}b_{3}-a_{3}b_{2}, a_{1}b_{3}-a_{3}b_{1},a_{1}b_{2}-a_{2}b_{1} \right \rangle$ .

Given $a=\left \langle 0,-3,3\right \rangle$ and $b=\left \langle -3,0,3\right \rangle$ , the cross product $a \times b$ is:

$\begin{Bmatrix} i&j &k \\ 0&-3 &3 \\ -3&0 &3 \end{Bmatrix}$

$=\left \langle (-3)(3)-(3)(0), (0)(3)-(3)(-3), (0)(0)-(-3)(-3) \right \rangle$

$=\left \langle -9-0, 0+9, 0-9 \right \rangle$

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

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Example Question #59 : Vector Calculations

What is the dot product of $a=\left \langle 3,5,7\right \rangle$ and $b=\left \langle -4,-1,2\right \rangle$ ?

Possible Answers:

Correct answer:

Explanation:

The dot product of two vectors is the sum of the products of the vectors' corresponding elements. Given $a=\left \langle 3,5,7\right \rangle$ and $b=\left \langle -4,-1,2\right \rangle$ , then:

$\left \langle 3,5,7\right \rangle\times \left \langle -4,-1,2\right \rangle$

$=(3\times -4)+(5\times -1)+(7\times 2)$

=(-12)+(-5)+(14)

=-17+14

=-3

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Example Question #60 : Vector Calculations

What is the dot product of $a=\left \langle 2,0,-2\right \rangle$ and $b=\left \langle 6,1,7\right \rangle$ ?

Possible Answers:

Correct answer:

Explanation:

The dot product of two vectors is the sum of the products of the vectors' corresponding elements. Given $a=\left \langle 2,0,-2\right \rangle$ and $b=\left \langle 6,1,7\right \rangle$ then:

$\left \langle 2,0,-2\right \rangle\times \left \langle 6,1,7\right \rangle$

$=(2\times 6)+(0\times 1)+(-2\times7)$

=12+0-14

=-2

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Calculus 2 : Vector Calculations

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #51 : Vector Calculations

Example Question #1062 : Calculus Ii

Example Question #53 : Vector Calculations

Example Question #54 : Vector Calculations

Example Question #55 : Vector Calculations

Example Question #56 : Vector Calculations

Example Question #57 : Vector Calculations

Example Question #58 : Vector Calculations

Example Question #59 : Vector Calculations

Example Question #60 : Vector Calculations

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