Calculus 2 : Parametric, Polar, and Vector

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #242 : Vector

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

Possible Answers:

\(\displaystyle 35\)

\(\displaystyle 38\)

\(\displaystyle 36\)

\(\displaystyle 37\)

\(\displaystyle 39\)

Correct answer:

\(\displaystyle 38\)

Explanation:

The dot product of two vectors is the sum of the products of the vectors' corresponding elements. Given \(\displaystyle a=\left \langle 9,1,2\right \rangle\) and \(\displaystyle b=\left \langle 3,1,5\right \rangle\), then:

\(\displaystyle \left \langle 9,1,2\right \rangle \times \left \langle 3,1,5\right \rangle\)

\(\displaystyle =(9\times3)+(1\times1)+(2\times5)\)

\(\displaystyle =27+1+10\)

\(\displaystyle =27+11\)

\(\displaystyle =38\)

Example Question #591 : Parametric, Polar, And Vector

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

Possible Answers:

\(\displaystyle 1\)

\(\displaystyle -2\)

\(\displaystyle -1\)

\(\displaystyle 2\)

\(\displaystyle 0\)

Correct answer:

\(\displaystyle 0\)

Explanation:

The dot product of two vectors is the sum of the products of the vectors' corresponding elements. Given \(\displaystyle a=\left \langle -1,-1,1\right \rangle\) and \(\displaystyle b=\left \langle 2,-1,1\right \rangle\),then:

\(\displaystyle \left \langle -1,-1,1\right \rangle \times \left \langle 2,-1,1\right \rangle\)

\(\displaystyle =(-1)(2)+(-1)(-1)+(1)(1)\)

\(\displaystyle =-2+1+1\)

\(\displaystyle =-2+2\)

\(\displaystyle =0\)

 

Example Question #83 : Vector Calculations

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

Possible Answers:

\(\displaystyle 12\)

\(\displaystyle 4\sqrt{33}\)

\(\displaystyle 11\)

\(\displaystyle \sqrt{133}\)

\(\displaystyle \sqrt{134}\)

Correct answer:

\(\displaystyle \sqrt{134}\)

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 \(\displaystyle a=\left \langle 2,-7,9\right \rangle\), then:

\(\displaystyle \left \| a\right \|=\sqrt{(2)^{2}+(-7)^{2}+(9)^{2}}\)

\(\displaystyle \left \| a\right \|=\sqrt{4+49+81}\)

\(\displaystyle \left \| a\right \|=\sqrt{53+81}\)

\(\displaystyle \left \| a\right \|=\sqrt{134}\)

Example Question #84 : Vector Calculations

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

Possible Answers:

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

\(\displaystyle 4\)

\(\displaystyle \sqrt{21}\)

\(\displaystyle 4\sqrt{5}\)

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

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

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

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

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

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

Example Question #246 : Vector

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

Possible Answers:

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

\(\displaystyle 7\sqrt{2}\)

\(\displaystyle 4\sqrt{2}\)

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

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

Correct answer:

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

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 \(\displaystyle a=\left \langle 4,3,5\right \rangle\), then:

\(\displaystyle \left \| a\right \|=\sqrt{(4)^{2}+(3)^{2}+(5)^{2}}\)

\(\displaystyle \left \| a\right \|=\sqrt{16+9+25}\)

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

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

\(\displaystyle \left \| a\right \|=\sqrt{25\times2}\)

\(\displaystyle \left \| a\right \|=5\sqrt{2}\)

 

Example Question #1108 : Calculus Ii

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

Possible Answers:

\(\displaystyle -21\)

\(\displaystyle 22\)

\(\displaystyle 20\)

\(\displaystyle 21\)

\(\displaystyle -22\)

Correct answer:

\(\displaystyle 22\)

Explanation:

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

\(\displaystyle \left \langle -8,4,3\right \rangle \times \left \langle -4,2,-6\right \rangle\)

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

\(\displaystyle =32+8-18\)

\(\displaystyle =32-10\)

\(\displaystyle =22\)

Example Question #251 : Vector

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

Possible Answers:

\(\displaystyle \left \langle -10,20,11 \right \rangle\)

\(\displaystyle \left \langle -10,-20,-11 \right \rangle\)

\(\displaystyle \left \langle 10,20,11 \right \rangle\)

\(\displaystyle \left \langle -10,-20,11 \right \rangle\)

\(\displaystyle \left \langle 10,-20,11 \right \rangle\)

Correct answer:

\(\displaystyle \left \langle -10,20,11 \right \rangle\)

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 and , then

.

Given  \(\displaystyle a=\left \langle 2,-1,0\right \rangle\) and \(\displaystyle b=\left \langle 1,5,10\right \rangle\), the cross product is:

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

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

\(\displaystyle =\left \langle -10-0,20-0,10+1 \right \rangle\)

\(\displaystyle =\left \langle -10,20,11 \right \rangle\)

 

 

Example Question #591 : Parametric, Polar, And Vector

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

Possible Answers:

\(\displaystyle \left \langle -42, -41,16 \right \rangle\)

\(\displaystyle \left \langle 42, 41,-16 \right \rangle\)

\(\displaystyle \left \langle 42, -41,-16 \right \rangle\)

\(\displaystyle \left \langle -42, 41,16 \right \rangle\)

\(\displaystyle \left \langle 42, -41,16 \right \rangle\)

Correct answer:

\(\displaystyle \left \langle 42, 41,-16 \right \rangle\)

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

\(\displaystyle \begin{Bmatrix} i& j&k \\ 5&2 &8 \\ -2&-4 &5 \end{Bmatrix}\)

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

\(\displaystyle =\left \langle 10+32, 25+16,-20+4 \right \rangle\)

\(\displaystyle =\left \langle 42, 41,-16 \right \rangle\)

 

Example Question #592 : Parametric, Polar, And Vector

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

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

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

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

 

Example Question #91 : Vector Calculations

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

Calculate the magnitude, or \(\displaystyle \left \| \vec{v} \right \|\), of \(\displaystyle \vec{v}\)

Possible Answers:

\(\displaystyle \left \| \vec{v} \right \|=\sqrt{156}\)

\(\displaystyle \left \| \vec{v} \right \|=\sqrt{154}\)

\(\displaystyle \left \| \vec{v} \right \|=\sqrt{157}\)

\(\displaystyle \left \| \vec{v} \right \|=\sqrt{155}\)

Correct answer:

\(\displaystyle \left \| \vec{v} \right \|=\sqrt{155}\)

Explanation:

Calculating magnitude:

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

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

Note: Magnitude, length, and norm are synonymous.

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

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

\(\displaystyle \left \| \vec{v} \right \|=\sqrt{25+121+9}\)

\(\displaystyle \left \| \vec{v} \right \|=\sqrt{155}\)

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

is \(\displaystyle \sqrt{155}\)

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