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Calculus 2 : Vector Calculations

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Example Question #61 : Vector Calculations

What is the dot product of $a=\left \langle 0,5,1\right \rangle$ and $b=\left \langle 5,1,0\right \rangle$ ?

Possible Answers:

Correct answer:

Explanation:

The dot product of two vectors is the sum of the products of the vectors' corresponding elements. Given $a=\left \langle 0,5,1\right \rangle$ and $b=\left \langle 5,1,0\right \rangle$ , then:

$\left \langle 0,5,1\right \rangle\times\left \langle 5,1,0\right \rangle$

$=(0\times5)+(5\times1)+(1\times0)$

=0+5+0

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Example Question #62 : Vector Calculations

What is the norm of $a=\left \langle 7,2,1\right \rangle$ ?

Possible Answers:

$3\sqrt{6}$

$3\sqrt{10}$

$3\sqrt{7}$

$3\sqrt{3}$

$3\sqrt{5}$

Correct answer:

$3\sqrt{6}$

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 $a=\left \langle 7,2,1\right \rangle$ , then:

$\left \| a\right \|=\sqrt{(7)^{2}+(2)^{2}+(1)^{2}}$

$\left \| a\right \|=\sqrt{49+4+1}$

$\left \| a\right \|=\sqrt{49+5}$

$\left \| a\right \|=\sqrt{54}$

$\left \| a\right \|=\sqrt{9\times6}$

$\left \| a\right \|=3\sqrt{6}$

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Example Question #63 : Vector Calculations

What is the norm of $a=\left \langle -3,-4,7\right \rangle$ ?

Possible Answers:

$5\sqrt{3}$

$6\sqrt{2}$

$\sqrt{73}$

$-5\sqrt{3}$

$\sqrt{74}$

Correct answer:

$\sqrt{74}$

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 $a=\left \langle -3,-4,7\right \rangle$ , then:

$\left \| a\right \|=\sqrt{(-3)^{2}+(-4)^{2}+(7)^{2}}$

$\left \| a\right \|=\sqrt{9+16+49}$

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

$\left \| a\right \|=\sqrt{74}$

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Example Question #64 : Vector Calculations

What is the norm of $a=\left \langle 0,1,-1\right \rangle$ ?

Possible Answers:

$2\sqrt{2}$

None of the above

$5\sqrt{2}$

$\sqrt{2}$

$3\sqrt{2}$

Correct answer:

$\sqrt{2}$

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 $a=\left \langle 0,1,-1\right \rangle$ , then:

$\left \| a\right \|=\sqrt{(0)^{2}+(1)^{2}+(-1)^{2}}$

$\left \| a\right \|=\sqrt{0+1+1}$

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

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Example Question #65 : Vector Calculations

What is the cross product of $a=\left \langle 5,2,9\right \rangle$ and $b=\left \langle -1,2,4\right \rangle$ ?

Possible Answers:

$\left \langle 10,-29,12 \right \rangle$

$\left \langle -10,-29,12 \right \rangle$

$\left \langle -10,-29,-12 \right \rangle$

$\left \langle 10,29,12 \right \rangle$

$\left \langle -10,29,12 \right \rangle$

Correct answer:

$\left \langle -10,29,12 \right \rangle$

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 $a=\left \langle 5,2,9\right \rangle$ and $b=\left \langle -1,2,4\right \rangle$ , the cross product $a \times b$ is:

$\begin{Bmatrix} i& j&k \\ 5&2 &9 \\ -1&2 &4 \end{Bmatrix}$

$=\left \langle (2)(4)-(9)(2),(5)(4)-(9)(-1),(5)(2)-(2)(-1) \right \rangle$

$=\left \langle 8-18,20+9,10+2 \right \rangle$

$=\left \langle -10,29,12 \right \rangle$

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Example Question #66 : Vector Calculations

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

Possible Answers:

$\left \langle 14,17,-23 \right \rangle$

$\left \langle 14,-17,-23 \right \rangle$

$\left \langle -14,-17,-23 \right \rangle$

$\left \langle 14,17,23 \right \rangle$

$\left \langle -14,17,23 \right \rangle$

Correct answer:

$\left \langle 14,17,-23 \right \rangle$

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

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

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

$=\left \langle 15-1,10+7,-2-21 \right \rangle$

$=\left \langle 14,17,-23 \right \rangle$

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Example Question #67 : Vector Calculations

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

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

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

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

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Example Question #68 : Vector Calculations

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

Possible Answers:

$2\sqrt{5}$

$\sqrt{22}$

$\sqrt{23}$

$\sqrt{21}$

$3\sqrt{5}$

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

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

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

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

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

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Example Question #69 : Vector Calculations

What is the norm of $a=\left \langle 3,2,6\right \rangle$ ?

Possible Answers:

$2\sqrt{2}$

$\sqrt{6}$

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 $a=\left \langle 3,2,6\right \rangle$ , then:

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

$\left \| a\right \|=\sqrt{9+4+36}$

$\left \| a\right \|=\sqrt{13+36}$

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

$\left \| a\right \|=7$

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Example Question #70 : Vector Calculations

What is the norm of $a=\left \langle 1,-1,5\right \rangle$ ?

Possible Answers:

$2\sqrt{3}$

$\sqrt{3}$

$4\sqrt{3}$

$5\sqrt{3}$

$3\sqrt{3}$

Correct answer:

$3\sqrt{3}$

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 $a=\left \langle 1,-1,5\right \rangle$ , then:

$\left \| a\right \|=\sqrt{(1)^{2}+(-1)^{2}+(5)^{2}}$

$\left \| a\right \|=\sqrt{1+1+25}$

$\left \| a\right \|=\sqrt{27}$

$\left \| a\right \|=\sqrt{9\times3}$

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

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Calculus 2 : Vector Calculations

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #61 : Vector Calculations

Example Question #62 : Vector Calculations

Example Question #63 : Vector Calculations

Example Question #64 : Vector Calculations

Example Question #65 : Vector Calculations

Example Question #66 : Vector Calculations

Example Question #67 : Vector Calculations

Example Question #68 : Vector Calculations

Example Question #69 : Vector Calculations

Example Question #70 : Vector Calculations

All Calculus 2 Resources

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