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Calculus 3 : 3-Dimensional Space

Study concepts, example questions & explanations for Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Equations Of Lines And Planes

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Remember the general equation of a line in vector form:

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

Lets apply this to our problem.

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

Distribute the

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

Now we simply do vector addition to get

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

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Example Question #111 : Calculus 3

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

Possible Answers:

82.1

32.712

None of the other answers.

42.192

52.388

Correct answer:

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 x,y,z .

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

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

The (acute) angle between any two vector is

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

Substituting, we have

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

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

$\theta \approx 42.192$ .

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Example Question #1 : Equations Of Lines And Planes

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

Possible Answers:

The line and the plane are parallel.

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

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

(-1,2,-1)

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

Correct answer:

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

Explanation:

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

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

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

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

Substituting this value of back into the components of the line gives us

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

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Example Question #1 : Equations Of Lines And Planes

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

Possible Answers:

42.667

47.333

Correct answer:

Explanation:

A quick way to notice the answer is 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 $\mathbf{n}_1 = <1,1,1>$

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

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

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

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Example Question #2 : Equations Of Lines And Planes

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

O=(0,0,0)

P=(5,4,0)

Q=(-3,-7,0)

Possible Answers:

y=0

x=0

z=0

5x-7y+z=4

Correct answer:

z=0

Explanation:

The equation of a plane is defined as

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

where is the normal vector of the plane.

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

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

and find their cross product.

The cross product is defined as the determinant of the matrix

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

Which is

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

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

=-23k

Which tells us the normal vector is

<0,0,-23>

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

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

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

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

Simplified gives the equation of the plane

z=0

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Example Question #1 : 3 Dimensional Space

Find the equation of the plane containing the points

O=(0,0,0) P=(0,1,5) Q=(0,-2,-4)

Possible Answers:

x=0

y-4z=1

z=5

y=0

Correct answer:

x=0

Explanation:

The equation of a plane is defined as

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

where is the normal vector of the plane.

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

OP=<0,1,5> and OQ=<0,-2,-4>

and find their cross product.

The cross product is defined as the determinant of the matrix

$\begin{vmatrix} i&j &k \\ 0&1 &5 \\ 0&-2 &-4 \end{vmatrix}$

Which is

$=det(\begin{vmatrix}1&5 \\ -2&-4 \end{vmatrix})i-det(\begin{vmatrix} 0&5 \\ 0&-4 \end{vmatrix})j+det(\begin{vmatrix} 0&1 \\ 0&-2 \end{vmatrix})k$

=(1*(-4)-(-2)*5)i-(0*(-4)-(0)*5)j+(0*(-2)-0*1)k

=6i

Which tells us the normal vector is

<6,0,0>

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

$<6,0,0>\bullet <x-0,y-0,z-0>=0$

$<6,0,0>\bullet <x,y,z>=0$

6x+0y+0z=0

Simplified gives the equation of the plane

x=0

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Example Question #3 : Equations Of Lines And Planes

Which of the following is an equation of a plane parallel to the plane x+4y-3z=10 ?

Possible Answers:

x+4y-6z=5

2x+4y-3z =5

None of the other answers

x+4y-3z =5

x+8y-3z=5

Correct answer:

x+4y-3z =5

Explanation:

Planes that are parallel to each other only differ (if at all) by the constant on the right-hand side (when both sides are simplified). Since x+4y-3z =5 has the same x,y,z coefficients as the given plane, they are parallel to each other.

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Example Question #2 : 3 Dimensional Space

Find a parametric representation of the curve of intersection of the cylinder 9x^2+y^2=9 and the plane x+y+z=7 .

Possible Answers:

$\vec{r}(t)= \left< \cos t, 3 \sin t, 7-\cos t - 3 \sin t\right>$

$\vec{r}(t)= \left< \cos t, 9 \sin t, 7-\cos t - 9 \sin t\right>$

$\vec{r}(t)= \left< \cos t, 3 \sin 3t, 7-\cos t - 3 \sin 3t\right>$

$\vec{r}(t)= \left< \cos t, \sin t, 7-\cos t - \sin t\right>$

Correct answer:

$\vec{r}(t)= \left< \cos t, 3 \sin t, 7-\cos t - 3 \sin t\right>$

Explanation:

We can begin by rewriting the expression for the cylinder as follows

$9x^2 + y^2 = 9 \Rightarrow x^2 + \left( \frac{y}{3} \right) ^2 = 1$ .

This tells us that $x = \cos t, y = 3 \sin t$ . Plugging this back into the equation for the plane x + y + z = 7 to find $z = 7 - x - y = 7 - \cos t - 3 \sin t$ .

This gives us the representation of the curve of intersection as

$\vec{r}(t)= \left< \cos t, 3 \sin t, 7-\cos t - 3 \sin t\right>$ .

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Example Question #1 : Equations Of Lines And Planes

Find the equation of the plane containing the following points

O=(2,3,4)

P=(1,5,3)

Q=(5,2,1)

Possible Answers:

x+5y+3z=0

5x-6y+7z=20

2x-2y+z=0

5x-3y+z=4

Correct answer:

5x-6y+7z=20

Explanation:

The equation of a plane is defined as

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

where

is the normal vector of the plane.

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

OP=<1-2,5-3,3-4>=<-1,-2,-1>

and

OQ=<5-2,2-3,1-4>=<3,-1,-3>

and find their cross product.

The cross product is defined as the determinant of the matrix

$\begin{vmatrix} i&j &k \\ -1&-2 &-1 \\ 3&-1 &-3 \end{vmatrix}$

Which is

$=det(\begin{vmatrix} -2&-1 \\ -1&-3 \end{vmatrix})i-det(\begin{vmatrix} -1&-1 \\ 3&-3 \end{vmatrix})j+det(\begin{vmatrix} -1&-2 \\ 3&-1 \end{vmatrix})k$

=(-2(-3)-(-1)*(-1))i-(-1(-3)-3(-1))j+(-1(-1)-3(-2))k

=5i-6j+7k

Which tells us the normal vector is

<5,-6,7>

Using the point

O=(2,3,4)

and the normal vector to find the equation of the plane yields

$<5,-6,7>\bullet<x-2,y-3,z-4>=0$

5(x-2)+(-6)(y-3)+7(z-4)=0

5x-10-6y+18+7z-28=0

Simplified gives the equation of the plane

5x-6y+7z=20

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Example Question #10 : Equations Of Lines And Planes

Find the angle in degrees between the planes x+5y-z = 0 and x+z =0

Possible Answers:

67.002

None of the other answers

74.207

23.669

14.282

Correct answer:

74.207

Explanation:

To find the angle between the planes, we find the angle between their normal vectors.

We have

$\mathbf{n}_1 = <1,5,-1>$ , for the first plane, and

$\mathbf{n}_2 = <1,0,-1>$ , for the second plane.

The angle between these two vectors is

$\theta = \cos^{-1}(\frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{\|\mathbf{n}_1\|\|\mathbf{n}_2\|})$

$= \cos^{-1}(\frac{<1,5,-1> \cdot <1,0,-1>}{\sqrt{1^2+(-5)^2+(-1)^2}\sqrt{1^2+0^2+(-1)^2}})$

$\approx \cos^{-1}(\frac{2}{7.34847}) \approx 74.207$

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Calculus 3 : 3-Dimensional Space

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1 : Equations Of Lines And Planes

Example Question #111 : Calculus 3

Example Question #1 : Equations Of Lines And Planes

Example Question #1 : Equations Of Lines And Planes

Example Question #2 : Equations Of Lines And Planes

Example Question #1 : 3 Dimensional Space

Example Question #3 : Equations Of Lines And Planes

Example Question #2 : 3 Dimensional Space

Example Question #1 : Equations Of Lines And Planes

Example Question #10 : Equations Of Lines And Planes

All Calculus 3 Resources

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