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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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

Example Questions

Example Question #202 : Vector

What is the norm of the vector $a=\left \langle 0,1,2\right \rangle$ ?

Possible Answers:

$\sqrt{3}$

$\sqrt{5}$

Correct answer:

$\sqrt{5}$

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 $a=\left \langle 0,1,2\right \rangle$ , then:

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

$\left \| a\right \|=\sqrt{0+1+4}$

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

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Example Question #203 : Vector

What is the norm of $a=\left \langle 3,-1,7\right \rangle$ ?

Possible Answers:

$2\sqrt{15}$

$\sqrt{61}$

$\sqrt{59}$

$\sqrt{65}$

Correct answer:

$\sqrt{59}$

Explanation:

In order to find the norm of a vector, we must first find the sum of the squares of the vector's individual elements and then take the square root of that sum. Given $a=\left \langle 3,-1,7\right \rangle$ ,

$\left \| a\right \|=\sqrt{(3)^{2}+(-1)^{2}+(7)^{2}}$

$\left \| a\right \|=\sqrt{9+1+49}$

$\left \| a\right \|=\sqrt{10+49}$

$\left \| a\right \|=\sqrt{59}$

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Example Question #204 : Vector

What is the norm of $a=\left \langle 2,0,3\right \rangle$ ?

Possible Answers:

$\sqrt{11}$

$\sqrt{13}$

$\sqrt{14}$

$2\sqrt{3}$

Correct answer:

$\sqrt{13}$

Explanation:

In order to find the norm of a vector, we must first find the sum of the squares of the vector's individual elements and then take the square root of that sum. Given $a=\left \langle 2,0,3\right \rangle$ ,

$\left \| a\right \|=\sqrt{(2)^{2}+(0)^{2}+(3)^{2}}$

$\left \| a\right \|=\sqrt{4+0+9}$

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

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Example Question #205 : Vector

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

Possible Answers:

$\sqrt{51}$

$5\sqrt{2}$

$4\sqrt{3}$

$2\sqrt{13}$

Correct answer:

$\sqrt{51}$

Explanation:

In order to find the norm of a vector, we must first find the sum of the squares of the vector's individual elements and then take the square root of that sum. Given $a=\left \langle 5,-5,1\right \rangle$ ,

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

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

$\left \| a\right \|=\sqrt{50+1}$

$\left \| a\right \|=\sqrt{51}$

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Example Question #47 : Vector Calculations

Find the cross product of $a=\left \langle 2,3,4\right \rangle$ and $b=\left \langle 3,2,1\right \rangle$ .

Possible Answers:

$\left \langle 5,-10,-5\right \rangle$

$\left \langle 5,10,-5\right \rangle$

None of the above

$\left \langle -5,-10,-5\right \rangle$

$\left \langle 5,10,-5\right \rangle$

Correct answer:

$\left \langle -5,-10,-5\right \rangle$

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 $a=\left \langle 2,3,4\right \rangle$ and $b=\left \langle 3,2,1\right \rangle$ , the cross product $a \times b$ is:

$\begin{Bmatrix} i&j &k \\ 2&3 &4 \\ 3&2 &1 \end{Bmatrix}$

$=\left \langle (3)(1)-(4)(2), (2)(1)-(4)(3),(2)(2)-(3)(3) \right \rangle$

$=\left \langle 3-8,2-12, 4-9\right \rangle$

$=\left \langle -5,-10,-5\right \rangle$

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Example Question #206 : Vector

Find the cross product of $a=\left \langle 1,0,-1\right \rangle$ and $b=\left \langle 0,-1,1\right \rangle$ .

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

In order to find the cross product of two 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 $a=\left \langle 1,0,-1\right \rangle$ and $b=\left \langle 0,-1,1\right \rangle$ . the cross product $a \times b$ is:

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

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

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

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

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Example Question #207 : Vector

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

Possible Answers:

$\left \langle8, 48, -8 \right \rangle$

$\left \langle8, 48, 8 \right \rangle$

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

$\left \langle8, -48, -8 \right \rangle$

Correct answer:

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

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

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

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

$=\left \langle(-7)-1, (-49)-(-1), (-7)-1 \right \rangle$

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

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Example Question #208 : Vector

What is the cross product of $a=\left \langle 5,6,-2\right \rangle$ and $b=\left \langle -3,1, 9 \right \rangle$ ?

Possible Answers:

$\left \langle 56, 39, 23 \right \rangle$

$\left \langle -56, -39, -23 \right \rangle$

$\left \langle -56, 39, -23 \right \rangle$

$\left \langle -56, 39, 23 \right \rangle$

$\left \langle -56, -39, 23 \right \rangle$

Correct answer:

$\left \langle 56, 39, 23 \right \rangle$

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 $a=\left \langle 5,6,-2\right \rangle$ and $b=\left \langle -3,1, 9 \right \rangle$ , the cross product $a \times b$ is:

$\begin{Bmatrix} i&j &k \\ 5&6 &-2 \\ -3&1 &9 \end{Bmatrix}$

$=\left \langle (6)(9)-(-2)(1), (5)(9)-(-2)(-3), (5)(1)-(6)(-3) \right \rangle$

$=\left \langle 54+2, 45-6, 5+18 \right \rangle$

$=\left \langle 56, 39, 23 \right \rangle$

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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:

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

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 #211 : Vector

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

Possible Answers:

$2\sqrt{15}$

$2\sqrt{14}$

$3\sqrt{6}$

$3\sqrt{7}$

$\sqrt{55}$

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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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #202 : Vector

Example Question #203 : Vector

Example Question #204 : Vector

Example Question #205 : Vector

Example Question #47 : Vector Calculations

Example Question #206 : Vector

Example Question #207 : Vector

Example Question #208 : Vector

Example Question #51 : Vector Calculations

Example Question #211 : Vector

All Calculus 2 Resources

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