Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #41 : Distance Between Vectors

\displaystyle \begin{align*}&\text{Find the distance between the two vectors:}\\&\overrightarrow{u}=\begin{bmatrix}8\\1\\-2\end{bmatrix}\overrightarrow{v}=\begin{bmatrix}-3\\10\\-16\end{bmatrix}\end{align*}

Possible Answers:

\displaystyle 19.95

\displaystyle 16.00

\displaystyle 21.68

\displaystyle 4.00

Correct answer:

\displaystyle 19.95

Explanation:

\displaystyle \begin{align*}&\text{The distance between two vectors, }\overrightarrow{u}\text{ and }\overrightarrow{v}\text{ is found}\\&\text{by summing the squares of the distances between paired coordinates and}\\&\text{taking the square root of this sum:}\\&\sqrt{\sum_{i=1}^{n}(\overrightarrow{u}_{i}-\overrightarrow{v}_{i})^2}\\&\text{It is similar to finding the diagonal length of a rectangle or rectangular prism.}\\&\text{For our vectors }\begin{bmatrix}8\\1\\-2\end{bmatrix}\text{ and }\begin{bmatrix}-3\\10\\-16\end{bmatrix}\text{ this becomes:}\\&\sqrt{[8-(-3)]^2+[1-(10)]^2+[-2-(-16)]^2}\\&\sqrt{(11)^2+(-9)^2+(14)^2}\\&19.95\end{align*}

Example Question #42 : Distance Between Vectors

\displaystyle \begin{align*}&\text{Find the distance between the two vectors:}\\&\overrightarrow{u}=\begin{bmatrix}-18\\8\\0\end{bmatrix}\overrightarrow{v}=\begin{bmatrix}-2\\5\\-17\end{bmatrix}\end{align*}

Possible Answers:

\displaystyle 29.29

\displaystyle 2.00

\displaystyle 23.54

\displaystyle 4.00

Correct answer:

\displaystyle 23.54

Explanation:

\displaystyle \begin{align*}&\text{The distance between two vectors, }\overrightarrow{u}\text{ and }\overrightarrow{v}\text{ is found}\\&\text{by summing the squares of the distances between paired coordinates and}\\&\text{taking the square root of this sum:}\\&\sqrt{\sum_{i=1}^{n}(\overrightarrow{u}_{i}-\overrightarrow{v}_{i})^2}\\&\text{It is similar to finding the diagonal length of a rectangle or rectangular prism.}\\&\text{For our vectors }\begin{bmatrix}-18\\8\\0\end{bmatrix}\text{ and }\begin{bmatrix}-2\\5\\-17\end{bmatrix}\text{ this becomes:}\\&\sqrt{[-18-(-2)]^2+[8-(5)]^2+[0-(-17)]^2}\\&\sqrt{(-16)^2+(3)^2+(17)^2}\\&23.54\end{align*}

Example Question #43 : Distance Between Vectors

\displaystyle \begin{align*}&\text{Calculate the distance between the two vectors:}\\&\begin{bmatrix}-1\\-20\\7\end{bmatrix}\text{ and }\begin{bmatrix}-13\\14\\6\end{bmatrix}\end{align*}

Possible Answers:

\displaystyle 20.02

\displaystyle 21.00

\displaystyle 36.07

\displaystyle 4.58

Correct answer:

\displaystyle 36.07

Explanation:

\displaystyle \begin{align*}&\text{The distance between two vectors, }\overrightarrow{u}\text{ and }\overrightarrow{v}\text{ is found}\\&\text{by summing the squares of the distances between paired coordinates and}\\&\text{taking the square root of this sum:}\\&\sqrt{\sum_{i=1}^{n}(\overrightarrow{u}_{i}-\overrightarrow{v}_{i})^2}\\&\text{It is similar to finding the diagonal length of a rectangle or rectangular prism.}\\&\text{For our vectors }\begin{bmatrix}-1\\-20\\7\end{bmatrix}\text{ and }\begin{bmatrix}-13\\14\\6\end{bmatrix}\text{ this becomes:}\\&\sqrt{[-1-(-13)]^2+[-20-(14)]^2+[7-(6)]^2}\\&\sqrt{(12)^2+(-34)^2+(1)^2}\\&36.07\end{align*}

Example Question #44 : Distance Between Vectors

Find the distance between the vectors \displaystyle \left \langle 3,-3,-2\right \rangle and \displaystyle \left \langle 1, 1, 7\right \rangle

Possible Answers:

\displaystyle \sqrt{101}

\displaystyle 10

\displaystyle \sqrt{111}

\displaystyle \sqrt{105}

Correct answer:

\displaystyle \sqrt{101}

Explanation:

To find the distance between two vectors \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, you use the formula:

\displaystyle d=\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+(a_3-b_3)^2}

Using the vectors from the problem statement, we get

\displaystyle d=\sqrt{(3-1)^2+(-3-1)^2+(-2-7)^2}=\sqrt{101}

Example Question #45 : Distance Between Vectors

Find the distance between the vectors \displaystyle \left \langle -5,-1,-3\right \rangle and \displaystyle \left \langle 2, 6, -4\right \rangle

Possible Answers:

\displaystyle \sqrt{99}

\displaystyle \sqrt{103}

\displaystyle 10

\displaystyle \sqrt{98}

Correct answer:

\displaystyle \sqrt{99}

Explanation:

To find the distance between two vectors \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, you use the formula:

\displaystyle d=\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+(a_3-b_3)^2}

Using the vectors from the problem statement, we get

\displaystyle d=\sqrt{(-5-2)^2+(-1-6)^2+(-3+4)^2}=\sqrt{99}

Example Question #111 : Calculus 3

Write down the equation of the line in vector form that passes through the points \displaystyle (2, 10 , -6), and \displaystyle (-3, 4, 14).

Possible Answers:

\displaystyle \vec{r}=\left \langle 2-5t, 10+6t, -6-20t\right \rangle

\displaystyle \vec{r}=\left \langle 2+5t, 10+6t, -6-20t\right \rangle

\displaystyle \vec{r}=\left \langle 2+5t, 1+t, -6-20t\right \rangle

\displaystyle \vec{r}=\left \langle 2+5t, 10-6t, -6-20t\right \rangle

\displaystyle \vec{r}=\left \langle 2+5t, 10+6t, -6-2t\right \rangle

Correct answer:

\displaystyle \vec{r}=\left \langle 2+5t, 10+6t, -6-20t\right \rangle

Explanation:

Remember the general equation of a line in vector form:

\displaystyle \vec{r}=r_0+t\vec{v}=\left \langle x_0,y_0,z_0\right \rangle+t\left \langle a,b,c\right \rangle, where \displaystyle \left \langle x_0,y_0,z_0\right \rangle is the starting point, and \displaystyle \left \langle a,b,c\right \rangle is the difference between the start and ending points.

Lets apply this to our problem.

\displaystyle \vec{r}=\left \langle 2, 10, -6\right \rangle+t\left \langle 5, 6, -20\right \rangle

Distribute the \displaystyle t

\displaystyle \vec{r}=\left \langle 2, 10, -6\right \rangle+\left \langle 5t, 6t, -20t\right \rangle

Now we simply do vector addition to get

\displaystyle \vec{r}=\left \langle 2+5t, 10+6t, -6-20t\right \rangle

Example Question #2 : Equations Of Lines And Planes

Find the approximate angle between the planes \displaystyle 4x-3y-2z=1, and \displaystyle 12x+2y-7z=16.

Possible Answers:

\displaystyle 82.1

\displaystyle 52.388

\displaystyle 32.712

\displaystyle 42.192

None of the other answers.

Correct answer:

\displaystyle 42.192

Explanation:

Finding the angle between two planes requires us to find the angle between their normal vectors.

To obtain normal vectors, we simply take the coefficients in front of \displaystyle x,y,z.

\displaystyle \mathbf{n_1} = < 4,-3,-2>

\displaystyle \mathbf{n_2} = < 12,2,-7>

The (acute) angle between any two vector is

\displaystyle \theta = \cos^{-1}(\frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|}),

Substituting, we have

\displaystyle \theta = \cos^{-1}(\frac{< 4,-3,-2> \cdot < 12,2,-7>}{\sqrt{16+9+4}\sqrt{144+4+49}})

\displaystyle \theta = \cos^{-1}(\frac{< 4,-3,-2> \cdot < 12,2,-7>}{\sqrt{16+9+4}\sqrt{144+4+49}})

\displaystyle \theta \approx \cos^{-1}(\frac{56}{75.584})

\displaystyle \theta \approx 42.192.

Example Question #1 : 3 Dimensional Space

Find the point of intersection of the plane \displaystyle 2x+y+z=9 and the line described by \displaystyle < 2t+4,t-1,-t>

Possible Answers:

\displaystyle (0,0,\frac{1}{2})

\displaystyle (5,-\frac{1}{2}, -\frac{1}{2})

\displaystyle (-1,2,-1)

The line and the plane are parallel.

\displaystyle (-\frac{1}{2},5,\frac{1}{2})

Correct answer:

\displaystyle (5,-\frac{1}{2}, -\frac{1}{2})

Explanation:

Substituting the components of the line into those of the plane, we have

\displaystyle 2(2t+4)+(t-1)+(-t)=9

\displaystyle 4t+8+t-1-t=9

\displaystyle t = \frac{1}{2}

Substituting this value of \displaystyle t back into the components of the line gives us

\displaystyle (5,-\frac{1}{2},-\frac{1}{2}).

Example Question #2 : 3 Dimensional Space

Find the angle (in degrees) between the planes \displaystyle x+y+z =2, \displaystyle x+y+z = 0

Possible Answers:

\displaystyle 47.333

\displaystyle 0

\displaystyle 42.667

\displaystyle 90

\displaystyle 30

Correct answer:

\displaystyle 0

Explanation:

A quick way to notice the answer is \displaystyle 0 is to notice the planes are parallel (They only differ by the constant on the right side).

Typically though, to find the angle between two planes, we find the angle between their normal vectors.

A vector normal to the first plane is \displaystyle \mathbf{n}_1 = < 1,1,1>

A vector normal to the second plane is \displaystyle \mathbf{n}_2 = < 1,1,1>

Then using the formula for the angle between vectors, \displaystyle \theta = \cos^{-1} (\frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}), we have

\displaystyle \theta = \cos^{-1} (\frac{< 1,1,1> \cdot < 1,1,1>}{\sqrt{1^2+1^2+1^2} \sqrt{1^2+1^2+1^2}}) = \cos^{-1}(1) = 0.

Example Question #2 : 3 Dimensional Space

Determine the equation of the plane that contains the following points. 

\displaystyle O=(0,0,0)

\displaystyle P=(5,4,0)

\displaystyle Q=(-3,-7,0)

Possible Answers:

\displaystyle 5x-7y+z=4

\displaystyle z=0

\displaystyle y=0

\displaystyle x=0

Correct answer:

\displaystyle z=0

Explanation:

The equation of a plane is defined as

\displaystyle n\bullet< x-x_0,y-y_0,z-z_0>=0

where \displaystyle n is the normal vector of the plane. 

To find the normal vector, we first get two vectors on the plane 

\displaystyle OP=< 5,4,0> and \displaystyle OQ=< -3,-7,0>

and find their cross product. 

The cross product is defined as the determinant of the matrix

\displaystyle \begin{vmatrix} i&j &k \\ 5&4 &0 \\ -3&-7 &0 \end{vmatrix}

Which is

\displaystyle =det(\begin{vmatrix} 4&0 \\ -7&0 \end{vmatrix})i-det(\begin{vmatrix} 5&0 \\ -3&0 \end{vmatrix})j+det(\begin{vmatrix} 5&4 \\ -3&-7 \end{vmatrix})k

\displaystyle =(4*0-(-7)*0)i-(5*0-(-3)*0)j+(5*(-7)-(-3)*4)k

\displaystyle =-23k

Which tells us the normal vector is 

\displaystyle < 0,0,-23>

Using the point \displaystyle O and the normal vector to find the equation of the plane yields

\displaystyle < 0,0,-23>\bullet < x-0,y-0,z-0>=0

\displaystyle < 0,0,-23>\bullet < x,y,z>=0

\displaystyle 0x+0y+(-23)z=0

Simplified gives the equation of the plane 

\displaystyle z=0

 

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