Calculus 2 : Vector Calculations

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #561 : Parametric, Polar, And Vector

What is the dot product of \(\displaystyle a=\left \langle 0,5,1\right \rangle\) and \(\displaystyle b=\left \langle 5,1,0\right \rangle\)?

Possible Answers:

\(\displaystyle 0\)

\(\displaystyle 1\)

\(\displaystyle 5\)

\(\displaystyle -5\)

\(\displaystyle -1\)

Correct answer:

\(\displaystyle 5\)

Explanation:

The dot product of two vectors is the sum of the products of the vectors' corresponding elements. Given \(\displaystyle a=\left \langle 0,5,1\right \rangle\) and \(\displaystyle b=\left \langle 5,1,0\right \rangle\), then:

\(\displaystyle \left \langle 0,5,1\right \rangle\times\left \langle 5,1,0\right \rangle\)

\(\displaystyle =(0\times5)+(5\times1)+(1\times0)\)

\(\displaystyle =0+5+0\)

\(\displaystyle =5\)

Example Question #221 : Vector

What is the norm of \(\displaystyle a=\left \langle 7,2,1\right \rangle\)?

Possible Answers:

\(\displaystyle 3\sqrt{7}\)

\(\displaystyle 3\sqrt{10}\)

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

\(\displaystyle 3\sqrt{6}\)

\(\displaystyle 3\sqrt{5}\)

Correct answer:

\(\displaystyle 3\sqrt{6}\)

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 \(\displaystyle a=\left \langle 7,2,1\right \rangle\) , then:

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

\(\displaystyle \left \| a\right \|=\sqrt{49+4+1}\)

\(\displaystyle \left \| a\right \|=\sqrt{49+5}\)

\(\displaystyle \left \| a\right \|=\sqrt{54}\)

\(\displaystyle \left \| a\right \|=\sqrt{9\times6}\)

\(\displaystyle \left \| a\right \|=3\sqrt{6}\)

Example Question #63 : Vector Calculations

What is the norm of \(\displaystyle a=\left \langle -3,-4,7\right \rangle\)?

Possible Answers:

\(\displaystyle \sqrt{73}\)

\(\displaystyle \sqrt{74}\)

\(\displaystyle -5\sqrt{3}\)

\(\displaystyle 6\sqrt{2}\)

\(\displaystyle 5\sqrt{3}\)

Correct answer:

\(\displaystyle \sqrt{74}\)

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 \(\displaystyle a=\left \langle -3,-4,7\right \rangle\) , then:

\(\displaystyle \left \| a\right \|=\sqrt{(-3)^{2}+(-4)^{2}+(7)^{2}}\)

\(\displaystyle \left \| a\right \|=\sqrt{9+16+49}\)

\(\displaystyle \left \| a\right \|=\sqrt{25+49}\)

\(\displaystyle \left \| a\right \|=\sqrt{74}\)

Example Question #64 : Vector Calculations

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

Possible Answers:

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

\(\displaystyle \sqrt{2}\)

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

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

None of the above

Correct answer:

\(\displaystyle \sqrt{2}\)

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

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

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

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

\(\displaystyle \left \| a\right \|=\sqrt{2}\)

 

Example Question #222 : Vector

What is the cross product of \(\displaystyle a=\left \langle 5,2,9\right \rangle\) and  \(\displaystyle b=\left \langle -1,2,4\right \rangle\)?

Possible Answers:

\(\displaystyle \left \langle -10,-29,12 \right \rangle\)

\(\displaystyle \left \langle -10,-29,-12 \right \rangle\)

\(\displaystyle \left \langle 10,-29,12 \right \rangle\)

\(\displaystyle \left \langle 10,29,12 \right \rangle\)

\(\displaystyle \left \langle -10,29,12 \right \rangle\)

Correct answer:

\(\displaystyle \left \langle -10,29,12 \right \rangle\)

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  \(\displaystyle a=\left \langle 5,2,9\right \rangle\) and  \(\displaystyle b=\left \langle -1,2,4\right \rangle\), the cross product is:

\(\displaystyle \begin{Bmatrix} i& j&k \\ 5&2 &9 \\ -1&2 &4 \end{Bmatrix}\)

 \(\displaystyle =\left \langle (2)(4)-(9)(2),(5)(4)-(9)(-1),(5)(2)-(2)(-1) \right \rangle\)

\(\displaystyle =\left \langle 8-18,20+9,10+2 \right \rangle\)

\(\displaystyle =\left \langle -10,29,12 \right \rangle\)

Example Question #571 : Parametric, Polar, And Vector

What is the cross product of \(\displaystyle a=\left \langle 2,3,-1\right \rangle\) and  \(\displaystyle b=\left \langle 7,-1,5\right \rangle\)?

Possible Answers:

\(\displaystyle \left \langle -14,17,23 \right \rangle\)

\(\displaystyle \left \langle -14,-17,-23 \right \rangle\)

\(\displaystyle \left \langle 14,-17,-23 \right \rangle\)

\(\displaystyle \left \langle 14,17,23 \right \rangle\)

\(\displaystyle \left \langle 14,17,-23 \right \rangle\)

Correct answer:

\(\displaystyle \left \langle 14,17,-23 \right \rangle\)

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  \(\displaystyle a=\left \langle 2,3,-1\right \rangle\) and  \(\displaystyle b=\left \langle 7,-1,5\right \rangle\), the cross product  is:

\(\displaystyle \begin{Bmatrix} i& j&k \\ 2&3 &-1 \\ 7&-1 &5 \end{Bmatrix}\)

\(\displaystyle =\left \langle (3)(5)-(-1)(-1),(2)(5)-(-1)(7), (2)(-1)-(3)(7) \right \rangle\)

\(\displaystyle =\left \langle 15-1,10+7,-2-21 \right \rangle\)

\(\displaystyle =\left \langle 14,17,-23 \right \rangle\)

 

 

Example Question #572 : Parametric, Polar, And Vector

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

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 three-dimensional vectors, we must find the determinant of a matrix comprised of the vectors' elements. That is, if  and , then

.

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

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

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

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

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

Example Question #571 : Parametric, Polar, And Vector

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

Possible Answers:

\(\displaystyle 3\sqrt{5}\)

\(\displaystyle \sqrt{21}\)

\(\displaystyle \sqrt{23}\)

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

\(\displaystyle \sqrt{22}\)

Correct answer:

\(\displaystyle \sqrt{21}\)

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 \(\displaystyle a=\left \langle 2,-1,4\right \rangle\), then:

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

\(\displaystyle \left \| a\right \|=\sqrt{4+1+16}\)

\(\displaystyle \left \| a\right \|=\sqrt{5+16}\)

\(\displaystyle \left \| a\right \|=\sqrt{21}\)

Example Question #67 : Vector Calculations

What is the norm of \(\displaystyle a=\left \langle 3,2,6\right \rangle\)?

Possible Answers:

\(\displaystyle 7\)

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

\(\displaystyle 10\)

\(\displaystyle 3\)

\(\displaystyle \sqrt{6}\)

Correct answer:

\(\displaystyle 7\)

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 \(\displaystyle a=\left \langle 3,2,6\right \rangle\), then:

\(\displaystyle \left \| a\right \|=\sqrt{(3)^{2}+(2)^{2}+(6)^{2}}\)

\(\displaystyle \left \| a\right \|=\sqrt{9+4+36}\)

\(\displaystyle \left \| a\right \|=\sqrt{13+36}\)

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

\(\displaystyle \left \| a\right \|=7\)

Example Question #223 : Vector

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

Possible Answers:

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

\(\displaystyle 5\sqrt{3}\)

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

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

\(\displaystyle \sqrt{3}\)

Correct answer:

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

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

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

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

\(\displaystyle \left \| a\right \|=\sqrt{27}\)

\(\displaystyle \left \| a\right \|=\sqrt{9\times3}\)

\(\displaystyle \left \| a\right \|=3\sqrt{3}\)

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