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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 #232 : Vector

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

Possible Answers:

$\sqrt{11}$

$\sqrt{13}$

$\sqrt{10}$

$3\sqrt{3}$

Correct answer:

$\sqrt{11}$

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 $a=\left \langle 1,-1,3\right \rangle$ , then:

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

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

$\left \| a\right \|=\sqrt{11}$

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Example Question #1092 : Calculus Ii

What is the cross product of $a=\left \langle 2,6,8\right \rangle$ and $b=\left \langle 1,-1,3\right \rangle$ ?

Possible Answers:

$\left \langle -26,-2, -8 \right \rangle$

$\left \langle 26,-2,8 \right \rangle$

$\left \langle 26,-2,-8 \right \rangle$

$\left \langle 26,-2, -8 \right \rangle$

$\left \langle 26,2,8 \right \rangle$

Correct answer:

$\left \langle 26,-2, -8 \right \rangle$

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 $a=\left \langle 2,6,8\right \rangle$ and $b=\left \langle 1,-1,3\right \rangle$ , the cross product $a \times b$ is:

$\begin{Bmatrix} i&j &k \\ 2&6 &8 \\ 1&-1 &3 \end{Bmatrix}$

$=\left \langle (6)(3)-(8)(-1),(2)(3)-(8)(1), (2)(-1)-(6)(1) \right \rangle$

$=\left \langle 18+8,6-8, -2-6 \right \rangle$

$=\left \langle 26,-2, -8 \right \rangle$

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Example Question #1091 : Calculus Ii

What is the cross product of $a=\left \langle 3,-4,2\right \rangle$ and $b=\left \langle 5,-10,2\right \rangle$ ?

Possible Answers:

$\left \langle 12,4,10 \right \rangle$

$\left \langle -12,-4,-10 \right \rangle$

$\left \langle 12,4,-10 \right \rangle$

$\left \langle 12,-4,-10 \right \rangle$

$\left \langle 12,-4,10 \right \rangle$

Correct answer:

$\left \langle 12,-4,-10 \right \rangle$

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 $a=\left \langle 3,-4,2\right \rangle$ and $b=\left \langle 5,-10,2\right \rangle$ the cross product $a \times b$ is:

$\begin{Bmatrix} i&j &k \\ 3&-4 &2 \\ 5&-10 &2 \end{Bmatrix}$

$=\left \langle (-4)(2)-(2)(-10),(3)(2)-(2)(5),(3)(-10)-(-4)(5) \right \rangle$

$=\left \langle -8+20,6-10,-30+20 \right \rangle$

$=\left \langle 12,-4,-10 \right \rangle$

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Example Question #1094 : Calculus Ii

What is the cross product of $a=\left \langle 0,1,1\right \rangle$ and $b=\left \langle 1,0,-4\right \rangle$ ?

Possible Answers:

$\left \langle 4,-1,1 \right \rangle$

$\left \langle -4,-1,1 \right \rangle$

$\left \langle 4,1,1 \right \rangle$

$\left \langle -4,-1,-1 \right \rangle$

$\left \langle -4,1,1 \right \rangle$

Correct answer:

$\left \langle -4,-1,-1 \right \rangle$

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 $a=\left \langle 0,1,1\right \rangle$ and $b=\left \langle 1,0,-4\right \rangle$ , the cross product $a \times b$ is:

$\begin{Bmatrix} i&j &k \\ 0&1 &1 \\ 1&0 &-4 \end{Bmatrix}$

$=\left \langle (1)(-4)-(1)(0),(0)(-4)-(1)(1),(0)(0)-(1)(1) \right \rangle$

$=\left \langle -4-0,0-1,0-1 \right \rangle$

$=\left \langle -4,-1,-1 \right \rangle$

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Example Question #233 : Vector

What is the dot product of $a=\left \langle 2,4,-6\right \rangle$ and $b=\left \langle -3,5,7\right \rangle$ ?

Possible Answers:

Correct answer:

Explanation:

The dot product of two vectors is the sum of the products of the vectors' corresponding elements. Given $a=\left \langle 2,4,-6\right \rangle$ and $b=\left \langle -3,5,7\right \rangle$ , then:

$a\times b=(2\times-3)+(4\times5)+(-6\times7)$

$a\times b=(-6)+20+(-42)$

$a\times b=(-48)+20$

$a\times b=-28$

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Example Question #234 : Vector

What is the dot product of $a=\left \langle 5,-4,8\right \rangle$ and $b=\left \langle 2,1,7\right \rangle$ ?

Possible Answers:

Correct answer:

Explanation:

The dot product of two vectors is the sum of the products of the vectors' corresponding elements. Given $a=\left \langle 5,-4,8\right \rangle$ and $b=\left \langle 2,1,7\right \rangle$ , then:

$a\times b=(5\times2)+(-4\times1)+(8\times7)$

$a\times b=10-4+56$

$a\times b=6+56$

$a\times b=62$

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Example Question #235 : Vector

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

Possible Answers:

Correct answer:

Explanation:

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

$a\times b=(1\times1)+(-1\times5)+(-5)(-1)$

$a\times b=1+(-5)+5$

$a\times b=1$

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Example Question #236 : Vector

What is the norm of $a=\left \langle 2,-4,5\right \rangle$ ?

Possible Answers:

$\sqrt{5}$

$2\sqrt{5}$

$3\sqrt{5}$

$4\sqrt{5}$

$5\sqrt{5}$

Correct answer:

$3\sqrt{5}$

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 $a=\left \langle 2,-4,5\right \rangle$ , then:

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

$\left \| a\right \|=\sqrt{4+16+25}$

$\left \| a\right \|=\sqrt{20+25}$

$\left \| a\right \|=\sqrt{45}$

$\left \| a\right \|=\sqrt{9\times5}$

$\left \| a\right \|=3\sqrt{5}$

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Example Question #237 : Vector

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

Possible Answers:

$5\sqrt{69}$

$2\sqrt{69}$

$4\sqrt{69}$

$\sqrt{69}$

$3\sqrt{69}$

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 #241 : Vector

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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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #232 : Vector

Example Question #1092 : Calculus Ii

Example Question #1091 : Calculus Ii

Example Question #1094 : Calculus Ii

Example Question #233 : Vector

Example Question #234 : Vector

Example Question #235 : Vector

Example Question #236 : Vector

Example Question #237 : Vector

Example Question #241 : Vector

All Calculus 2 Resources

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