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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 #31 : Equations Of Lines And Planes

Find the equation of the plane given by a point on the plane $\left ( 2,-1,-4\right )$ and the normal vector to the plane $n=\left \langle 2,3,9\right \rangle$

Possible Answers:

2x+3y+8z=34

2x+3y-9z=36

2x+3y+9z=-35

2x+3y+9z=35

Correct answer:

2x+3y+9z=-35

Explanation:

To find the equation of the plane, we use the formula A(x-x_0)+B(y-y_0)+C(z-z_0)=0 , where the point given is (x_0,y_0,z_0) and the normal vector $n=\left \langle A,B,C\right \rangle$ . Plugging in what we were given in the problem statement, we get 2(x-2)+3(y+1)+9(z+4)=0 . Manipulating the equation through algebra to isolate the variables, we get 2x+3y+9z=-35

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Example Question #141 : Calculus 3

Find the equation of the plane given by a point on the plane $\left ( -1,2,-4\right )$ and the normal vector to the plane $n=\left \langle 0,4,5\right \rangle$

Possible Answers:

x+4y+5z=12

x+5z=12

4y+5z=12

4y+5z=-12

Correct answer:

4y+5z=-12

Explanation:

To find the equation of the plane, we use the formula A(x-x_0)+B(y-y_0)+C(z-z_0)=0 , where the point given is (x_0,y_0,z_0) and the normal vector $n=\left \langle A,B,C\right \rangle$ . Plugging in what we were given in the problem statement, we get 0(x+1)+4(y-2)+5(z+4)=0 . Manipulating the equation through algebra to isolate the variables, we get 4y+5z=-12

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Example Question #31 : 3 Dimensional Space

Find the equation of the line that passes through the points (0,2,1) and (5,4,3)

Note: The answer you obtain needs to be in vector form. Also, use the point (0,2,1) when forming the equation

Possible Answers:

$\left \langle 4t,3t+1,t\right \rangle$

$\left \langle 4t,t+1,-2t+2\right \rangle$

$\left \langle 3t,2t+1,3\right \rangle$

$\left \langle -5t,-2t+2,-2t+1\right \rangle$

Correct answer:

$\left \langle -5t,-2t+2,-2t+1\right \rangle$

Explanation:

To find the equation of the line, we need to find the vector that will be parallel to the line (its direction). This vector is formed by the points a=(x_1,y_1,z_1) and b=(x_2,y_2,z_2) (the ones from the problem statement.) We find the direction vector to be $\left \langle x_1-x_2,y_1-y_2,z_1-z_2\right \rangle=\left \langle 0-5,2-4,1-3\right \rangle=\left \langle -5,-2,-2\right \rangle$

We then pick a point on the line. We chose (0,2,1) . We then form the equation of the line by using the formula $\left \langle 0,1,2\right \rangle+t\left \langle-5,-2,-2 \right \rangle=\left \langle -5t,-2t+2,2t+1\right \rangle$

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Example Question #145 : Calculus 3

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

Possible Answers:

2x+3y+z=10

3x+2y-9z=20

x+2y+5z=-9

x+4y+2z=8

Correct answer:

2x+3y+z=10

Explanation:

To find the equation of the plane, we use the formula A(x-x_0)+B(y-y_0)+C(z-z_0)=0 , where the point given is (x_0,y_0,z_0) and the normal vector $n=\left \langle A,B,C\right \rangle$ . Plugging in what we were given in the problem statement, we get 2(x-5)+3(y+1)+1(z-3)=0 . Manipulating the equation through algebra to isolate the variables, we get 2x+3y+z=10

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Example Question #144 : Calculus 3

Find the equation of the plane containing the point $\left ( 3,-2,1\right )$ , and is parallel to the plane with the equation 2x+5y+z=20

Possible Answers:

x+2y+5z=4

2x+5y+z=14

2x+5y+z=-3

2x+5y+z=3

Correct answer:

2x+5y+z=-3

Explanation:

We were given a point on the plane, and we need the normal vector to the plane. It is known that two planes that are parallel to each other have the same normal vector, so in this case $n=\left \langle 2,5,1\right \rangle$ (given by the equation of the other plane). To complete the problem, we use the equation A(x-x_0)+B(y-y_0)+C(z-z_0)=0 , where $n=\left \langle A,B,C\right \rangle$ and the point on the plane is (x_0,y_0,z_0) . Using the information we have, we get:

2(x-3)+5(y+2)+1(z-1)=0 . Through algebraic manipulation, we then get:

2x+5y+z=-3

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Example Question #151 : Calculus 3

Find the equation of the plane containing the point $\left ( 4,0,2\right )$ , and is parallel to the plane with the equation 5x+2y+2z=15

Possible Answers:

5x+2y+z=24

5x+2y+2z=16

5x+2y+2z=21

5x+2y+2z=24

Correct answer:

5x+2y+2z=24

Explanation:

We were given a point on the plane, and we need the normal vector to the plane. It is known that two planes that are parallel to each other have the same normal vector, so in this case $n=\left \langle 5,2,2\right \rangle$ (given by the equation of the other plane). To complete the problem, we use the equation A(x-x_0)+B(y-y_0)+C(z-z_0)=0 , where $n=\left \langle A,B,C\right \rangle$ and the point on the plane is (x_0,y_0,z_0) . Using the information we have, we get:

5(x-4)+2(y-0)+2(z-2)=0 . Through algebraic manipulation, we then get:

5x+2y+2z=24

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Example Question #32 : Equations Of Lines And Planes

Find the equation of the plane that contains the point (5,-1,2) and is parallel to the plane 2x+y+3z=6

Possible Answers:

2x+y+3z=12

2x+y+3z=15

2x+y=11

2x+y-3z=15

Correct answer:

2x+y+3z=15

Explanation:

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

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

We have the point, which from the problem statement is (5,-1,2) .

The normal vector is given by the equation of the plane that is in parallel to the one we are forming the equation for. That is $n=\left \langle 2,1,3\right \rangle$ .

Putting everything we know into the formula and solving, we get

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

Simplifying, we get

2x+y+3z=15

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Example Question #33 : Equations Of Lines And Planes

Find the equation of the plane that contains the point (6,11,10) and is parallel to the plane 3x+-y+2z=5

Possible Answers:

x-6y+2z=15

3x-y+2z=49

3x+y+2z=49

x-y+5z=9

Correct answer:

3x-y+2z=49

Explanation:

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

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

We have the point, which from the problem statement is (6,11,10) .

The normal vector is given by the equation of the plane that is in parallel to the one we are forming the equation for. That is $n=\left \langle 3,-1,2\right \rangle$ .

Putting everything we know into the formula and solving, we get

3(x-6)-1(y+11)+2(z-10)=0

Simplifying, we get

3x-y+2z=49

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Example Question #34 : Equations Of Lines And Planes

Find the equation of the plane given by a point on the plane (-1,3,-4) and the normal vector to the plane $n=\left \langle 6,7,1\right \rangle$

Possible Answers:

6x+7y-z=21

6x+7y+z=11

6x-7y+z=11

6x+7y-z=-11

Correct answer:

6x+7y+z=11

Explanation:

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

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

Using the information from the problem statement, we then get

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

Rearranging through algebra, we get:

6x+7y+z=11

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Example Question #35 : Equations Of Lines And Planes

Find the equation of the plane that contains the point (3,-1,4) and is parallel to the plane 4x-8y+z=22

Possible Answers:

4x-8y+z=24

4x+8y+z=12

4x-8y+z=15

4x-8y+z=20

Correct answer:

4x-8y+z=24

Explanation:

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

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

To find the normal vector, it is known that two parallel planes have the same normal vector. Using this and the point on the plane, we then get

4(x-3)+-8(y+1)+1(z-4)=0

Rearranging through algebra, we get:

4x-8y+z=24

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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 #31 : Equations Of Lines And Planes

Example Question #141 : Calculus 3

Example Question #31 : 3 Dimensional Space

Example Question #145 : Calculus 3

Example Question #144 : Calculus 3

Example Question #151 : Calculus 3

Example Question #32 : Equations Of Lines And Planes

Example Question #33 : Equations Of Lines And Planes

Example Question #34 : Equations Of Lines And Planes

Example Question #35 : Equations Of Lines And Planes

All Calculus 3 Resources

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