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Calculus 2 : Vector Calculations

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Example Question #91 :Vector Calculations

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 #252 :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 #253 :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 #254 :Vector

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

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

Possible Answers:

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

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

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

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

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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Example Question #255 :Vector

Vector $\vec{v} = (-6,6,9)$ .

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

Possible Answers:

$\left \| \vec{v} \right \|=9$

$\left \| \vec{v} \right \|=\sqrt{153}$

$\left \| \vec{v} \right \|=\sqrt{150}$

$\left \| \vec{v} \right \|=\sqrt{151}$

Correct answer:

$\left \| \vec{v} \right \|=\sqrt{153}$

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.

$\vec{v} = (-6,6,9)$

$\left \| \vec{v} \right \|=\sqrt{(-6)^2+6^2+9^2}$

$\left \| \vec{v} \right \|=\sqrt{36+36+81}$

$\left \| \vec{v} \right \|=\sqrt{153}$

$\left \| \vec{v} \right \|$ cannot be further reduced, and so $\left \| \vec{v} \right \|=\sqrt{153}$

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Example Question #256 :Vector

Vector $\vec{v} = (sin(x),cos(x),-1)$ .

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

Possible Answers:

$\left \| \vec{v} \right \|=1$

$\left \| \vec{v} \right \|=0$

$\left \| \vec{v} \right \|=2$

$\left \| \vec{v} \right \|=\sqrt{2}$

Correct answer:

$\left \| \vec{v} \right \|=\sqrt{2}$

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.

Note: sin(x)^2 + cos(x)^2 = 1 , and this is a basic trig identity that you should know.

$\vec{v} = (sin(x),cos(x),-1)$

$\left \| \vec{v} \right \|=\sqrt{sin(x)^2+cos(x)^2+(-1)^2}$

$\left \| \vec{v} \right \|=\sqrt{sin(x)^2+cos(x)^2+1}$

$\left \| \vec{v} \right \|=\sqrt{1 + 1}$

$\left \| \vec{v} \right \|=\sqrt{2}$

$\left \| \vec{v} \right \|$ cannot be further reduced, and so $\left \| \vec{v} \right \|=\sqrt{2}$

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Example Question #257 :Vector

Vector $\vec{v} = (2x^3,3x^{3},\sqrt{12}x^3)$ .

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

Possible Answers:

$\left \| \vec{v} \right \|=4x^3$

$\left \| \vec{v} \right \|=5x^3$

$\left \| \vec{v} \right \|=7x^3$

$\left \| \vec{v} \right \|=6x^3$

Correct answer:

$\left \| \vec{v} \right \|=5x^3$

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.

$\vec{v} = (2x^3,3x^{3},\sqrt{12}x^3)$

$\left \| \vec{v} \right \|=\sqrt{(2x^3)^2+(3x^3)^2+(\sqrt{12}x^3)^2}$

$\left \| \vec{v} \right \|=\sqrt{4x^6+9x^6+12x^6}$

$\left \| \vec{v} \right \|=\sqrt{25x^6}$

$\left \| \vec{v} \right \|=5x^3$

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Example Question #6 :Vector Form

$\newline\vec{v} =(x^2,-x,5)\newline\vec{u} =(2,3x,11)\newline$

Calculate $\vec{v} + \vec{u}$

Possible Answers:

$\vec{v}+\vec{u}=(x^2,2x,16)$

$\vec{v}+\vec{u}=(x^2,2x,3x)$

$\vec{v}+\vec{u}=(x^2 + 2,0,16)$

$\vec{v}+\vec{u}=(x^2 + 2,2x,16)$

Correct answer:

$\vec{v}+\vec{u}=(x^2 + 2,2x,16)$

Explanation:

Calculate the sum of vectors.

In general,

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

$\vec{u} = (u_1,u_2,u_3)$

$\vec{v} + \vec{u} = (v_1 + u_1, v_2 + u_2,v_3 + u_3)$

Solution:

$\newline\vec{v} =(x^2,-x,5)\newline\vec{u} =(2,3x,11)\newline$

$\vec{v}+\vec{u}=(x^2,-x,5)+(2,3x,11)$

$\vec{v}+\vec{u}=(x^2+2,-x+3x,5+11)$

$\vec{v}+\vec{u}=(x^2 + 2,2x,16)$

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Example Question #258 :Vector

$\newline\vec{v} =(2x+3,3x^2+1,e^x+2)\newline\vec{u} =(-2x+4,-11x^2+2,7e^x+x^2+3)\newline$

Calculate $\vec{v} + \vec{u}$

Possible Answers:

$\vec{v}+\vec{u}=(7,-8x^2+3,8e^x+x^2+5)$

$\vec{v}+\vec{u}=(7,-8x+3,8e^x+x^2+4)$

$\vec{v}+\vec{u}=(7,-8x^2+3,e^{8x}+x^2+5)$

$\vec{v}+\vec{u}=(7,-8x^2,8e^x+x^2+5)$

Correct answer:

$\vec{v}+\vec{u}=(7,-8x^2+3,8e^x+x^2+5)$

Explanation:

Calculate the sum of vectors.

In general,

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

$\vec{u} = (u_1,u_2,u_3)$

$\vec{v} + \vec{u} = (v_1 + u_1, v_2 + u_2,v_3 + u_3)$

Solution:

$\vec{v} =(2x+3,3x^2+1,e^x+2)$

$\vec{u} =(-2x+4,-11x^2+2,7e^x+x^2+3)$

$\vec{v}+\vec{u}=(2x+3,3x^2+1,e^x+2)+(-2x+4,-11x^2+2,7e^x+x^2+3)$

$\vec{v}+\vec{u}=(2x+3+-2x+4,3x^2+1+-11x^2+2,e^x+2+7e^x+x^2+3)$

$\vec{v}+\vec{u}=(7,-8x^2+3,8e^x+x^2+5)$

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Calculus 2 : Vector Calculations

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #91 :Vector Calculations

Example Question #251 :Vector

Example Question #252 :Vector

Example Question #253 :Vector

Example Question #254 :Vector

Example Question #255 :Vector

Example Question #256 :Vector

Example Question #257 :Vector

Example Question #6 :Vector Form

Example Question #258 :Vector

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