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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 #41 : 3 Dimensional Space

Find the equation of the plane given by the normal vector $n=\left \langle 7,5,2\right \rangle$ and a point on the plane $\left ( 1,3,5\right )$

Possible Answers:

7x+5y+2z=22

7x+5y-4z=10

7x+5y+2z=32

Correct answer:

7x+5y+2z=32

Explanation:

To find the equation of the plane with a normal vector $n=\left \langle A,B,C\right \rangle$ and a point (x_0,y_0,z_0) , we use the formula

A(x-x_0)+B(y-y_0)+C(z-z_0)=0

Using the information from the problem statement, we get

7(x-1)+5(y-3)+2(z-5)=0

This simplifies to 7x+5y+2z=32

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Example Question #41 : Equations Of Lines And Planes

Find the equation of the plane given by the normal vector $n=\left \langle 2,-5,6\right \rangle$ and a point on the plane $\left ( 3,7,9\right )$

Possible Answers:

2x-5y+6z=16

2x+5y+6z=25

2x-5y+6z=25

2x-5y+6z=24

Correct answer:

2x-5y+6z=25

Explanation:

To find the equation of the plane with a normal vector $n=\left \langle A,B,C\right \rangle$ and a point (x_0,y_0,z_0) , we use the formula

A(x-x_0)+B(y-y_0)+C(z-z_0)=0

Using the information from the problem statement, we get

2(x-3)-5(y-7)+6(z-9)=0

This simplifies to 2x-5y+6z=25

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Example Question #151 : Calculus 3

Find the equation of the plane that contains the point $\left ( 3,-6,7\right )$ and has a normal vector $n=\left \langle 4,2,1\right \rangle$

Possible Answers:

3x+2y+z=7

4x+2y+z=7

4x+2y+z=14

4x+2y-z=7

Correct answer:

4x+2y+z=7

Explanation:

To find the equation of a plane with a point (x_0,y_0,z_0) and a normal vector $n=\left \langle A,B,C\right \rangle$ , we use the formula

A(x-x_0)+B(y-y_0)+C(z-z_0)=0

Using the information from the problem statement, we get

4(x-3)+2(y+6)+1(z-7)=0

Simplifying, we then get

4x+2y+z=7

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Example Question #44 : 3 Dimensional Space

Find the equation of the plane that contains the point (5,-1,2) and contains the normal vector $\left \langle 4,5,5\right \rangle$

Possible Answers:

4x+5y+5z=20

2x+5y+5z=25

4x+5y+5z=25

4x+y+3z=25

Correct answer:

4x+5y+5z=25

Explanation:

To find the equation of a plane with a point (x_0,y_0,z_0) and a normal vector $n=\left \langle A,B,C\right \rangle$ , we use the formula

A(x-x_0)+B(y-y_0)+C(z-z_0)=0

Using the information from the problem statement, we get

4(x-5)+5(y+1)+5(z-2)=0

Simplifying, we then get

4x+5y+5z=25

Report an Error

Example Question #45 : 3 Dimensional Space

Find the equation of the plane containing these points:

A=(1, 2, 3) , B=(0, 3, -6) , C=(-2, 4, 2)

Possible Answers:

x+10y+z=-24

x+2y+3z=24

x+10y+z=0

x+10y+z=24

Correct answer:

x+10y+z=24

Explanation:

To find the equation of a plane, we need the normal vector and a point on the plane. The normal vector is found by taking the cross product of two vectors on the plane.

To find two vectors, simply find the difference between terminal and initial points:

$\vec{AB}=\left \langle -1, 1, -9\right \rangle$ , $\vec{BC}=\left \langle -2, 1, -8 \right \rangle$

Now, we can write the determinant in order to take the cross product of the two vectors:

$\begin{vmatrix} \hat{i}& \hat{j}&\hat{k} \\ -1&1 &-9 \\ -2&1 &-8 \end{vmatrix}$

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

Next, we take the cross product. One can do this by multiplying across from the top left to the lower right, and continuing downward, and then subtracting the terms multiplied from top right to the bottom left:

$-8\hat{i}-\hat{k}+18\hat{j}-(-2\hat{k}-9\hat{i}+8\hat{j})=\hat{i}+10\hat{j}+\hat{k}=\left \langle 1, 10, 1\right \rangle$

Plugging this into the formula

a(x-x_0)+b(y-y_0)+c(z-z_0)=0 , where $\vec{n}=\left \langle a, b, c\right \rangle$ and (x_0, y_0, z_0) is any of the given points (for example (1, 2 ,3) ), we get

1(x-1)+10(y-2)+1(z-3)=0

which simplified becomes

x+10y+z=24

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Example Question #161 : Calculus 3

Find the equation of the plane that contains the point $\left ( 3,0,1\right )$ and has a normal vector $n=\left \langle 4,4,1\right \rangle$

Possible Answers:

4x+3y+z=13

4x+4y+z=10

4x+4y+z=13

4x+4y-z=13

Correct answer:

4x+4y+z=13

Explanation:

To find the equation of a plane with a point (x_0,y_0,z_0) and normal vector $n=\left \langle A,B,C\right \rangle$ , we use the following equation:

A(x-x_0)+B(y-y_0)+C(z-z_0)=0

Plugging in the information from the problem statement, we get

4(x-3)+4(y-0)+1(z-1)=0

Isolating the variables to one side gets us

4x+4y+z=13

Report an Error

Example Question #41 : Equations Of Lines And Planes

Find the equation of the plane that contains the point $\left ( 2,1,5\right )$ and has a normal vector $n=\left \langle 2,5,2\right \rangle$

Possible Answers:

` 2x+6y+2z=19

2x-5y+2z=19

2x+5y+2z=14

2x+5y+2z=19

Correct answer:

2x+5y+2z=19

Explanation:

To find the equation of a plane with a point (x_0,y_0,z_0) and normal vector $n=\left \langle A,B,C\right \rangle$ , we use the following equation:

A(x-x_0)+B(y-y_0)+C(z-z_0)=0

Plugging in the information from the problem statement, we get

2(x-2)+5(y-1)+2(z-5)=0

Isolating the variables to one side gets us

2x+5y+2z=19

Report an Error

Example Question #41 : Equations Of Lines And Planes

Find the equation of the plane that contains the point (3,-6,7) and a normal vector $n=\left \langle 1,2,1\right \rangle$

Possible Answers:

x-2y-4z=-4

x+2y+z=-2

2x+2y+z=-2

x+2y-z=-2

Correct answer:

x+2y+z=-2

Explanation:

To find the equation of the plane containing a point (x_0,y_0,z_0) and a normal vector $n=\left \langle A,B,C\right \rangle$ , we use the formula:

A(x-x_0)+B(y-y_0)+C(z-z_0)=0

Plugging in the known values and solving, we get

1(x-3)+2(y+6)+1(z-7)=0

Simplifying, we get

x+2y+z=-2

Report an Error

Example Question #163 : Calculus 3

Find the equation of the plane that contains the point (5,4,1) and a normal vector $n=\left \langle 3,1,5\right \rangle$

Possible Answers:

3x+y+5z=24

5x+y+5z=22

3x-y+5z=24

3x+4y+5z=24

Correct answer:

3x+y+5z=24

Explanation:

To find the equation of the plane containing a point (x_0,y_0,z_0) and a normal vector $n=\left \langle A,B,C\right \rangle$ , we use the formula:

A(x-x_0)+B(y-y_0)+C(z-z_0)=0

Plugging in the known values and solving, we get

3(x-5)+1(y-4)+5(z-1)=0

Simplifying, we get

3x+y+5z=24

Report an Error

Example Question #162 : Calculus 3

Find the equation of the plane given by a point on the plane (3,-4,3) and the normal vector $n=\left \langle 8,4,-5\right \rangle$

Possible Answers:

8x+4y-5z=-7

8x+2y-5z=-7

6x+4y-5z=-7

8x+4y-3z=-7

Correct answer:

8x+4y-5z=-7

Explanation:

To find the equation of a plane given a point on the plane (x_0,y_0,z_0) and a normal vector to the plane $n=\left \langle A,B,C\right \rangle$ , we use the following equation

A(x-x_0)+B(y-y_0)+C(z-z_0)=0

Plugging in the information from the problem statement, we get

8(x-3)+4(y+4)-5(z-3)=0

Rearranging, we get

8x+4y-5z=-7

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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 #41 : 3 Dimensional Space

Example Question #41 : Equations Of Lines And Planes

Example Question #151 : Calculus 3

Example Question #44 : 3 Dimensional Space

Example Question #45 : 3 Dimensional Space

Example Question #161 : Calculus 3

Example Question #41 : Equations Of Lines And Planes

Example Question #41 : Equations Of Lines And Planes

Example Question #163 : Calculus 3

Example Question #162 : Calculus 3

All Calculus 3 Resources

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