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Calculus 2 : Vector Calculations

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Example Questions

Example Question #81 : Vector Calculations

What is the norm of $a=\left \langle 4,2,-7\right \rangle$ ?

Possible Answers:

$\sqrt{69}$

$4\sqrt{69}$

$2\sqrt{69}$

$5\sqrt{69}$

$3\sqrt{69}$

Correct answer:

$\sqrt{69}$

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 $a=\left \langle 4,2,-7\right \rangle$ , then:

$\left \| a\right \|=\sqrt{(4)^{2}+(2)^{2}+(-7)^{2}}$

$\left \| a\right \|=\sqrt{16+4+49}$

$\left \| a\right \|=\sqrt{20+49}$

$\left \| a\right \|=\sqrt{69}$

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Example Question #82 : Vector Calculations

What is the norm of $a=\left \langle 1,-2,2\right \rangle$ ?

Possible Answers:

Correct answer:

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 $a=\left \langle 1,-2,2\right \rangle$ , then:

$\left \| a\right \|=\sqrt{(1)^{2}+(-2)^{2}+(2)^{2}}$

$\left \| a\right \|=\sqrt{1+4+4}$

$\left \| a\right \|=\sqrt{1+8}$

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

$\left \| a\right \|=3$

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Example Question #83 : Vector Calculations

What is the cross product of $a=\left \langle 2,7,-3\right \rangle$ and $b=\left \langle 5,1,-6\right \rangle$ .

Possible Answers:

$\left \langle -39,3, -33 \right \rangle$

$\left \langle -39,-3, 33 \right \rangle$

$\left \langle 39,3, -33 \right \rangle$

$\left \langle -39,-3, -33 \right \rangle$

$\left \langle 3,-39, -33 \right \rangle$

Correct answer:

$\left \langle -39,3, -33 \right \rangle$

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 $a=\left \langle 2,7,-3\right \rangle$ and $b=\left \langle 5,1,-6\right \rangle$ , the cross product $a \times b$ is:

$\left \langle 2,7,-3\right \rangle \times \left \langle 5,1,-6\right \rangle$

$=\begin{Bmatrix} i& j&k \\ 2&7 &-3 \\ 5&1 &-6 \end{Bmatrix}$

$=\left \langle (7)(-6)-(-3)(1),(2)(-6)-(-3)(5), (2)(1)-(7)(5) \right \rangle$

$=\left \langle -42+3,-12+15, 2-35 \right \rangle$

$=\left \langle -39,3, -33 \right \rangle$

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Example Question #84 : Vector Calculations

What is the cross product of $a=\left \langle 4,-1,1\right \rangle$ and $b=\left \langle 3,5,9\right \rangle$ .

Possible Answers:

$\left \langle -14, -39,-17 \right \rangle$

$\left \langle -14, 39,-17 \right \rangle$

$\left \langle 14, 39,17 \right \rangle$

$\left \langle -14, 39,17 \right \rangle$

$\left \langle -14, -39,17 \right \rangle$

Correct answer:

$\left \langle 14, 39,17 \right \rangle$

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 $a=\left \langle 4,-1,1\right \rangle$ and $b=\left \langle 3,5,9\right \rangle$ , the cross product $a \times b$ is:

$\left \langle 4,1,-1\right \rangle \times \left \langle 3,5,9\right \rangle$

$=\begin{Bmatrix} i& j&k \\ 4&1 &-1 \\ 3&5 &9 \end{Bmatrix}$

$=\left \langle (1)(9)-(-1)(5), (4)(9)-(-1)(3),(4)(5)-(1)(3) \right \rangle$

$=\left \langle 9+5, 36+3,20-3 \right \rangle$

$=\left \langle 14, 39,17 \right \rangle$

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Example Question #85 : Vector Calculations

What is the cross product of $a=\left \langle 1,-2,1\right \rangle$ and $b=\left \langle -1,-1,1\right \rangle$ .

Possible Answers:

$\left \langle -1, 2, -3 \right \rangle$

$\left \langle 1, 2, 3 \right \rangle$

$\left \langle 1, 2, -3 \right \rangle$

$\left \langle -1, -2, -3 \right \rangle$

$\left \langle 1, 2, -3 \right \rangle$

Correct answer:

$\left \langle -1, 2, -3 \right \rangle$

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 $a=\left \langle 1,-2,1\right \rangle$ and $b=\left \langle -1,-1,1\right \rangle$ , the cross product $a \times b$ is:

$\left \langle 1,-2,1\right \rangle \times \left \langle -1,-1,1\right \rangle$

$=\begin{Bmatrix} i& j&k \\ 1&-2 &1 \\ -1&-1 &1 \end{Bmatrix}$

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

$=\left \langle -2+1, 1+1, -1-2 \right \rangle$

$=\left \langle -1, 2, -3 \right \rangle$

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Example Question #1101 : Calculus Ii

What is the dot product of $a=\left \langle 9,1,2\right \rangle$ and $b=\left \langle 3,1,5\right \rangle$ ?

Possible Answers:

Correct answer:

Explanation:

The dot product of two vectors is the sum of the products of the vectors' corresponding elements. Given $a=\left \langle 9,1,2\right \rangle$ and $b=\left \langle 3,1,5\right \rangle$ , then:

$\left \langle 9,1,2\right \rangle \times \left \langle 3,1,5\right \rangle$

$=(9\times3)+(1\times1)+(2\times5)$

=27+1+10

=27+11

=38

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Example Question #1102 : Calculus Ii

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

Possible Answers:

Correct answer:

Explanation:

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

$\left \langle -1,-1,1\right \rangle \times \left \langle 2,-1,1\right \rangle$

=(-1)(2)+(-1)(-1)+(1)(1)

=-2+1+1

=-2+2

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Example Question #1103 : Calculus Ii

What is the norm of $a=\left \langle 2,-7,9\right \rangle$ ?

Possible Answers:

$\sqrt{134}$

$4\sqrt{33}$

$\sqrt{133}$

Correct answer:

$\sqrt{134}$

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 $a=\left \langle 2,-7,9\right \rangle$ , then:

$\left \| a\right \|=\sqrt{(2)^{2}+(-7)^{2}+(9)^{2}}$

$\left \| a\right \|=\sqrt{4+49+81}$

$\left \| a\right \|=\sqrt{53+81}$

$\left \| a\right \|=\sqrt{134}$

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Example Question #1104 : Calculus Ii

What is the norm of $a=\left \langle -1,2,4\right \rangle$ ?

Possible Answers:

$4\sqrt{5}$

$\sqrt{21}$

$2\sqrt{3}$

$3\sqrt{2}$

Correct answer:

$\sqrt{21}$

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 $a=\left \langle -1,2,4\right \rangle$ , then:

$\left \| a\right \|=\sqrt{(-1)^{2}+(2)^{2}+(4)^{2}}$

$\left \| a\right \|=\sqrt{1+4+16}$

$\left \| a\right \|=\sqrt{5+16}$

$\left \| a\right \|=\sqrt{21}$

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Example Question #1105 : Calculus Ii

What is the norm of $a=\left \langle 4,3,5\right \rangle$ ?

Possible Answers:

$5\sqrt{2}$

$6\sqrt{2}$

$7\sqrt{2}$

$3\sqrt{2}$

$4\sqrt{2}$

Correct answer:

$5\sqrt{2}$

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 $a=\left \langle 4,3,5\right \rangle$ , then:

$\left \| a\right \|=\sqrt{(4)^{2}+(3)^{2}+(5)^{2}}$

$\left \| a\right \|=\sqrt{16+9+25}$

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

$\left \| a\right \|=\sqrt{50}$

$\left \| a\right \|=\sqrt{25\times2}$

$\left \| a\right \|=5\sqrt{2}$

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Calculus 2 : Vector Calculations

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #81 : Vector Calculations

Example Question #82 : Vector Calculations

Example Question #83 : Vector Calculations

Example Question #84 : Vector Calculations

Example Question #85 : Vector Calculations

Example Question #1101 : Calculus Ii

Example Question #1102 : Calculus Ii

Example Question #1103 : Calculus Ii

Example Question #1104 : Calculus Ii

Example Question #1105 : Calculus Ii

All Calculus 2 Resources

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