Calculus 2 : Vector Calculations

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #546 : Parametric, Polar, And Vector

What is the norm of the vector \(\displaystyle a=\left \langle 7,-4,1\right \rangle\)?

Possible Answers:

\(\displaystyle \sqrt{63}\)

\(\displaystyle \sqrt{67}\)

\(\displaystyle \sqrt{66}\)

\(\displaystyle 8\)

\(\displaystyle \sqrt{65}\)

Correct answer:

\(\displaystyle \sqrt{66}\)

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 \(\displaystyle a=\left \langle 7,-4,1\right \rangle\), then:

\(\displaystyle \left \| a\right \|=\sqrt{(7)^{2}+(-4)^{2}+(1)^{2}}\)

\(\displaystyle \left \| a\right \|=\sqrt{49+16+1}\)

\(\displaystyle \left \| a\right \|=\sqrt{65+1}\)

\(\displaystyle \left \| a\right \|=\sqrt{66}\)

 

Example Question #41 : Vector Calculations

What is the norm of the vector \(\displaystyle a=\left \langle 2,3,-1\right \rangle\)?

Possible Answers:

\(\displaystyle \sqrt{15}\)

\(\displaystyle 3\)

\(\displaystyle \sqrt{17}\)

\(\displaystyle \sqrt{14}\)

\(\displaystyle 4\)

Correct answer:

\(\displaystyle \sqrt{14}\)

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 \(\displaystyle a=\left \langle 2,3,-1\right \rangle\), then:

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

\(\displaystyle \left \| a\right \|=\sqrt{4+9+1}\)

\(\displaystyle \left \| a\right \|=\sqrt{13+1}\)

\(\displaystyle \left \| a\right \|=\sqrt{14}\)

 

Example Question #548 : Parametric, Polar, And Vector

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

Possible Answers:

\(\displaystyle \sqrt{3}\)

\(\displaystyle 0\)

\(\displaystyle 2\)

\(\displaystyle \sqrt{5}\)

\(\displaystyle 1\)

Correct answer:

\(\displaystyle \sqrt{5}\)

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 \(\displaystyle a=\left \langle 0,1,2\right \rangle\), then:

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

\(\displaystyle \left \| a\right \|=\sqrt{0+1+4}\)

\(\displaystyle \left \| a\right \|=\sqrt{5}\)

Example Question #549 : Parametric, Polar, And Vector

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

Possible Answers:

\(\displaystyle \sqrt{61}\)

\(\displaystyle 8\)

\(\displaystyle \sqrt{65}\)

\(\displaystyle 2\sqrt{15}\)

\(\displaystyle \sqrt{59}\)

Correct answer:

\(\displaystyle \sqrt{59}\)

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 \(\displaystyle a=\left \langle 3,-1,7\right \rangle\),

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

\(\displaystyle \left \| a\right \|=\sqrt{9+1+49}\)

\(\displaystyle \left \| a\right \|=\sqrt{10+49}\)

\(\displaystyle \left \| a\right \|=\sqrt{59}\)

Example Question #42 : Vector Calculations

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

Possible Answers:

\(\displaystyle \sqrt{13}\)

\(\displaystyle \sqrt{14}\)

\(\displaystyle \sqrt{11}\)

\(\displaystyle 3\)

\(\displaystyle 2\sqrt{3}\)

Correct answer:

\(\displaystyle \sqrt{13}\)

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 \(\displaystyle a=\left \langle 2,0,3\right \rangle\),

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

\(\displaystyle \left \| a\right \|=\sqrt{4+0+9}\)

\(\displaystyle \left \| a\right \|=\sqrt{13}\)

Example Question #41 : Vector Calculations

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

Possible Answers:

\(\displaystyle 5\sqrt{2}\)

\(\displaystyle 4\sqrt{3}\)

\(\displaystyle \sqrt{51}\)

\(\displaystyle 7\)

\(\displaystyle 2\sqrt{13}\)

Correct answer:

\(\displaystyle \sqrt{51}\)

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 \(\displaystyle a=\left \langle 5,-5,1\right \rangle\),

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

\(\displaystyle \left \| a\right \|=\sqrt{25+25+1}\)

\(\displaystyle \left \| a\right \|=\sqrt{50+1}\)

\(\displaystyle \left \| a\right \|=\sqrt{51}\)

Example Question #552 : Parametric, Polar, And Vector

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

Possible Answers:

\(\displaystyle \left \langle -5,-10,-5\right \rangle\)

\(\displaystyle \left \langle 5,10,-5\right \rangle\)

\(\displaystyle \left \langle 5,10,-5\right \rangle\)

None of the above

\(\displaystyle \left \langle 5,-10,-5\right \rangle\)

Correct answer:

\(\displaystyle \left \langle -5,-10,-5\right \rangle\)

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 \(\displaystyle a=\left \langle a_{1},a_{2},a_{3}\right \rangle\) and \(\displaystyle b=\left \langle b_{1},b_{2},b_{3}\right \rangle\), then

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

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

\(\displaystyle \begin{Bmatrix} i&j &k \\ 2&3 &4 \\ 3&2 &1 \end{Bmatrix}\)

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

\(\displaystyle =\left \langle 3-8,2-12, 4-9\right \rangle\)

\(\displaystyle =\left \langle -5,-10,-5\right \rangle\)

Example Question #553 : Parametric, Polar, And Vector

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

Possible Answers:

\(\displaystyle \left \langle -1, 1,1 \right \rangle\)

\(\displaystyle \left \langle -1, -1,-1 \right \rangle\)

\(\displaystyle \left \langle -1, 1,-1 \right \rangle\)

\(\displaystyle \left \langle 1, 1,1 \right \rangle\)

\(\displaystyle \left \langle 1, 1,1 \right \rangle\)

Correct answer:

\(\displaystyle \left \langle -1, 1,-1 \right \rangle\)

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 \(\displaystyle a=\left \langle a_{1},a_{2},a_{3}\right \rangle\) and \(\displaystyle b=\left \langle b_{1},b_{2},b_{3}\right \rangle\), then

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

Given \(\displaystyle a=\left \langle 1,0,-1\right \rangle\) and \(\displaystyle b=\left \langle 0,-1,1\right \rangle\). the cross product \(\displaystyle a \times b\) is:

\(\displaystyle \begin{Bmatrix} i&j &k \\ 1&0 &-1 \\ 0&-1 &1 \end{Bmatrix}\)

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

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

\(\displaystyle =\left \langle -1, 1,-1 \right \rangle\)

 

Example Question #551 : Parametric, Polar, And Vector

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

Possible Answers:

\(\displaystyle \left \langle8, -48, -8 \right \rangle\)

\(\displaystyle \left \langle8, 48, -8 \right \rangle\)

\(\displaystyle \left \langle8, 48, 8 \right \rangle\)

\(\displaystyle \left \langle-8, -48, -8 \right \rangle\)

\(\displaystyle \left \langle-8, -48, -8 \right \rangle\)

Correct answer:

\(\displaystyle \left \langle-8, -48, -8 \right \rangle\)

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 and , then

.

Given  \(\displaystyle a=\left \langle 7,1,-1\right \rangle\) and \(\displaystyle b=\left \langle 1,-1, -7 \right \rangle\) the cross product is:

\(\displaystyle \begin{Bmatrix} i&j &k \\ 7&1 &-1 \\ 1&-1 &-7 \end{Bmatrix}\)

 

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

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

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

\(\displaystyle =\left \langle-8, -48, -8 \right \rangle\)

Example Question #555 : Parametric, Polar, And Vector

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

Possible Answers:

\(\displaystyle \left \langle -56, -39, 23 \right \rangle\)

\(\displaystyle \left \langle 56, 39, 23 \right \rangle\)

\(\displaystyle \left \langle -56, 39, -23 \right \rangle\)

\(\displaystyle \left \langle -56, 39, 23 \right \rangle\)

\(\displaystyle \left \langle -56, -39, -23 \right \rangle\)

Correct answer:

\(\displaystyle \left \langle 56, 39, 23 \right \rangle\)

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  and , then

.

Given  \(\displaystyle a=\left \langle 5,6,-2\right \rangle\) and \(\displaystyle b=\left \langle -3,1, 9 \right \rangle\),  the cross product  is:

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

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

\(\displaystyle =\left \langle 54+2, 45-6, 5+18 \right \rangle\)

\(\displaystyle =\left \langle 56, 39, 23 \right \rangle\)

 

 

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