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Calculus 2 : Parametric, Polar, and Vector

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 #242 : Vector

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 #243 : Vector

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 #244 : Vector

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

Possible Answers:

$\sqrt{133}$

$4\sqrt{33}$

$\sqrt{134}$

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 #245 : Vector

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

Possible Answers:

$2\sqrt{3}$

$\sqrt{21}$

$4\sqrt{5}$

$3\sqrt{2}$

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 #246 : Vector

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

Possible Answers:

$5\sqrt{2}$

$7\sqrt{2}$

$4\sqrt{2}$

$6\sqrt{2}$

$3\sqrt{2}$

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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Example Question #1108 : Calculus Ii

What is the dot product of $a=\left \langle -8,4,3\right \rangle$ and $b=\left \langle -4,2,-6\right \rangle$ ?

Possible Answers:

Correct answer:

Explanation:

The dot product of two vectors is the sum of the products of the vectors' corresponding elements. Given $a=\left \langle -8,4,3\right \rangle$ and $b=\left \langle -4,2,-6\right \rangle$ , then:

$\left \langle -8,4,3\right \rangle \times \left \langle -4,2,-6\right \rangle$

$=(-8\times-4)+(4\times2)+(3\times-6)$

=32+8-18

=32-10

=22

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Example Question #251 : Vector

What is the cross product of $a=\left \langle 2,-1,0\right \rangle$ and $b=\left \langle 1,5,10\right \rangle$ ?

Possible Answers:

$\left \langle -10,20,11 \right \rangle$

$\left \langle -10,-20,-11 \right \rangle$

$\left \langle 10,20,11 \right \rangle$

$\left \langle -10,-20,11 \right \rangle$

$\left \langle 10,-20,11 \right \rangle$

Correct answer:

$\left \langle -10,20,11 \right \rangle$

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 $a=\left \langle 2,-1,0\right \rangle$ and $b=\left \langle 1,5,10\right \rangle$ , the cross product $a \times b$ is:

$\begin{Bmatrix} i& j&k \\ 2&-1 &0 \\ 1&5 &10 \end{Bmatrix}$

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

$=\left \langle -10-0,20-0,10+1 \right \rangle$

$=\left \langle -10,20,11 \right \rangle$

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Example Question #591 : Parametric, Polar, And Vector

What is the cross product of $a=\left \langle 5,2,8\right \rangle$ and $b=\left \langle -2,-4,5\right \rangle$ ?

Possible Answers:

$\left \langle 42, -41,-16 \right \rangle$

$\left \langle 42, 41,-16 \right \rangle$

$\left \langle 42, -41,16 \right \rangle$

$\left \langle -42, 41,16 \right \rangle$

$\left \langle -42, -41,16 \right \rangle$

Correct answer:

$\left \langle 42, 41,-16 \right \rangle$

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

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

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

$=\left \langle 10+32, 25+16,-20+4 \right \rangle$

$=\left \langle 42, 41,-16 \right \rangle$

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Example Question #592 : Parametric, Polar, And Vector

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

Possible Answers:

$\left \langle -1,-7,3 \right \rangle$

$\left \langle 1,7,-3 \right \rangle$

$\left \langle 1,-7,-3 \right \rangle$

$\left \langle 1,-7,3 \right \rangle$

$\left \langle -1,7,3 \right \rangle$

Correct answer:

$\left \langle 1,-7,3 \right \rangle$

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

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

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

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

$=\left \langle 1,-7,3 \right \rangle$

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Example Question #1111 : Calculus Ii

Vector $\vec{v} = (5,11,-3)$ .

Calculate the magnitude, or $\left \| \vec{v} \right \|$ , of $\vec{v}$

Possible Answers:

$\left \| \vec{v} \right \|=\sqrt{156}$

$\left \| \vec{v} \right \|=\sqrt{154}$

$\left \| \vec{v} \right \|=\sqrt{155}$

$\left \| \vec{v} \right \|=\sqrt{157}$

Correct answer:

$\left \| \vec{v} \right \|=\sqrt{155}$

Explanation:

Calculating magnitude:

If $\vec{v} = (v_1,v_2,v_3)$

Then the magnitude of $\vec{v}$ is $\sqrt{v_1^2 + v_2^2 + v_3^2}$

Note: Magnitude, length, and norm are synonymous.

$\vec{v} = (5,11,-3)$

$\left \| \vec{v} \right \|=\sqrt{5^2+11^2+(-3)^2}$

$\left \| \vec{v} \right \|=\sqrt{25+121+9}$

$\left \| \vec{v} \right \|=\sqrt{155}$

$\left \| \vec{v} \right \|$ cannot be further reduced, so the magnitude of $\vec{v}$

is $\sqrt{155}$

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Calculus 2 : Parametric, Polar, and Vector

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #242 : Vector

Example Question #243 : Vector

Example Question #244 : Vector

Example Question #245 : Vector

Example Question #246 : Vector

Example Question #1108 : Calculus Ii

Example Question #251 : Vector

Example Question #591 : Parametric, Polar, And Vector

Example Question #592 : Parametric, Polar, And Vector

Example Question #1111 : Calculus Ii

All Calculus 2 Resources

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