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Calculus 3 : Calculus 3

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

Example Questions

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

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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

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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

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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

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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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Example Question #162 : Calculus 3

Find the equation of the plane given by a point on the plane (-1,7,5) and the normal vector $n=\left \langle 7,4,2\right \rangle$

Possible Answers:

7x+4y+2z=31

7x+y+2z=24

5x+4y+4z=31

x+4y+z=15

Correct answer:

7x+4y+2z=31

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

7(x+1)+4(y-7)+2(z-5)=0

Rearranging, we get

7x+4y+2z=31

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Example Question #167 : Calculus 3

Find the equation of the plane that contains the point (3,6,2) and is parallel to the plane 3x-5y+z=15

Possible Answers:

4x-5y+2z=-10

x-5y+z=19

3x+5y+z=-19

3x-5y+z=-19

Correct answer:

3x-5y+z=-19

Explanation:

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

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

Since we know the point on the plane as well as the normal vector (two parallel planes contain the same normal vector, so in this case it is $\left \langle 3,-5,1\right \rangle$ , we can plug what we know into the equation

3(x-3)-5(y-6)+1(z-2)=0

Rearranging, we get

3x-5y+z=-19

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Example Question #168 : Calculus 3

Find the equation of the plane that contains the point (2,-2,1) and is parallel to the plane 4x+2y-6z=24

Possible Answers:

4x+2y-6z=12

4x+2y-6z=-2

4x+3y+6z=-2

4x+y-4z=-2

Correct answer:

4x+2y-6z=-2

Explanation:

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

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

Since we know the point on the plane as well as the normal vector (two parallel planes contain the same normal vector, so in this case it is $\left \langle 4,2,-6\right \rangle$ , we can plug what we know into the equation

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

Rearranging, we get

4x+2y-6z=-2

Report an Error

Example Question #51 : Equations Of Lines And Planes

Find the equation of the plane given by the following points:

A=(1, 4, 1) , B=(2, 2, 2) , C=(5, 1, 3)

Possible Answers:

2x+2y+2z=12

-x+2y+5z=12

x+4y+z=12

-x+2y+5z=0

Correct answer:

-x+2y+5z=12

Explanation:

The equation of a plane is given by

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 point on the plane.

First, we must create two vectors out of the given points (by subtracting terminal and initial points):

$\vec{AB}=\left \langle 1, -2, 1 \right \rangle$ , $\vec{BC}=\left \langle 3, -1, 1\right \rangle$

Now, we can write the determinant in order to take the cross product of the two vectors, which will give us the normal vector:

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

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:

$-2\hat{i}-\hat{k}+3\hat{j}-(-6\hat{k}-\hat{i}+\hat{j})= --1\hat{i}+2\hat{k}+5\hat{j}=\left \langle -1, 2, 5\right \rangle$

Now that we have the normal vector, we can pick any point on the plane, and plug all of this into the formula above:

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

which simplified becomes

-x+2y+5z=12

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Example Question #53 : 3 Dimensional Space

Determine the equation of the plane given by the following two vectors and the point (4, 2, 3) :

$\left \langle 3, 1, 5\right \rangle, \left \langle 2, 1, 1\right \rangle$

Possible Answers:

4x+2y+3z=0

4x+2y+3z=1

-4x+7y+z=-1

-4x+7y+z=0

-4x+7y+z=1

Correct answer:

-4x+7y+z=1

Explanation:

The equation of a plane is given by

a(x-x_0)+b(y-y_0)+c(z-z_0)=0

where the normal vector is given by $\vec{n}=\left \langle a,b,c\right \rangle$ and a point on the plane denoted (x_0, y_0, z_0)

To find the normal vector to the plane, we must take the cross product of the two vectors.

We must write the determinant in order to take the cross product:

$\begin{vmatrix} \hat{i}& \hat{j}&\hat{k} \\ 3&1 &5 \\ 2&1 &1 \end{vmatrix}$

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

$\hat{i}+3\hat{k}+10\hat{j}-(2\hat{k}+5\hat{i}+3\hat{j})= -4\hat{i}+7\hat{j}+\hat{k}=\left \langle -4, 7, 1\right \rangle$

Now that we have the normal vector and a point on the plane, we plug everything into the equation:

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

which simplifies to

-4x+7y+z=1

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

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

Example Question #162 : Calculus 3

Example Question #167 : Calculus 3

Example Question #168 : Calculus 3

Example Question #51 : Equations Of Lines And Planes

Example Question #53 : 3 Dimensional Space

All Calculus 3 Resources

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