Find the scalar, vector, and parametric equations of a plane

In summary: So this tells me that PR does not work. Now what can I do to fix this?You can try changing the direction vector for PR.
  • #1
Specter
120
8

Homework Statement


Find the scalar, vector, and parametric equations of a plane that has a normal vector n=(3,-4,6) and passes through point P(9,2,-5)

Homework Equations

The Attempt at a Solution


Finding the scalar equation:

Ax+By+Cz+D=0
3x-4y+6z+D=0
3(9)-4(2)+6(-5)+D=0
-11+D=0
D=11

3x-4y+6z+11=0 is the scalar equation.

To find the vector equation I need a point and two direction vectors.

3x-4y+6z+11=0
6z=-3x+4y-11
z=-3/6 x + 4/6 y -11/6
z=-1/2 x +4/6 y - 11/6

Let x=0 and y=0

z=-11/6

One point is Q(0,0,-11/6)

Let x=0 and z=1

3(0)-4y+6(1)+11=0
-4y+6+11=0
-4y+17=0
y=17/4

Another point is R(0, 17/4, 1)

Now I have 3 points, I need 2 direction vectors.

PQ=Q-P=(0,0,-11/6)-(9,2,-5)
=(-9,-2,19/6)

PR=R-P=(0,17/4,0)-(9,2,-5)
=(-9,9/4,5)

The vector equation should be (x,y,z)=(9,2,-5)+t(-9,-2,19/6)+s(-9,9/4,5), where s and t are any real numbers.

Parametric equations are

x=9-9t-9s
y=2-2t+9/4 s
z=-5+19/6 t + 5s

I think I did this correctly but I'm not sure how I can check to see if my answer is correct. If anyone could look this over and let me know how I could check it myself would be great. Thank you.
 
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  • #2
Specter said:

Homework Statement


Find the scalar, vector, and parametric equations of a plane that has a normal vector n=(3,-4,6) and passes through point P(9,2,-5)

Homework Equations

The Attempt at a Solution


Finding the scalar equation:

Ax+By+Cz+D=0
3x-4y+6z+D=0
3(9)-4(2)+6(-5)+D=0
-11+D=0
D=11

3x-4y+6z+11=0 is the scalar equation.
I get this as well.
Specter said:
To find the vector equation I need a point and two direction vectors.

3x-4y+6z+11=0
6z=-3x+4y-11
z=-3/6 x + 4/6 y -11/6
z=-1/2 x +4/6 y - 11/6

Let x=0 and y=0

z=-11/6

One point is Q(0,0,-11/6)

Let x=0 and z=1

3(0)-4y+6(1)+11=0
-4y+6+11=0
-4y+17=0
y=17/4

Another point is R(0, 17/4, 1)

Now I have 3 points, I need 2 direction vectors.

PQ=Q-P=(0,0,-11/6)-(9,2,-5)
=(-9,-2,19/6)

PR=R-P=(0,17/4,0)-(9,2,-5)
=(-9,9/4,5)
You have a mistake here.
Specter said:
The vector equation should be (x,y,z)=(9,2,-5)+t(-9,-2,19/6)+s(-9,9/4,5), where s and t are any real numbers.

Parametric equations are

x=9-9t-9s
y=2-2t+9/4 s
z=-5+19/6 t + 5s

I think I did this correctly but I'm not sure how I can check to see if my answer is correct. If anyone could look this over and let me know how I could check it myself would be great. Thank you.
You can use your scalar equation to find a couple more points (which you did).
Verify that the two direction vectors you found are actually perpendicular to the given normal vector. Note that one of them doesn't work.
You can also check your parametric equations by verifying that there are specific values of t and s that generate the points you already know.
 
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  • #3
Mark44 said:
I get this as well.
You have a mistake here.

You can use your scalar equation to find a couple more points (which you did).
Verify that the two direction vectors you found are actually perpendicular to the given normal vector. Note that one of them doesn't work.
You can also check your parametric equations by verifying that there are specific values of t and s that generate the points you already know.
How can I verify that the direction vectors are perpendicular to the normal vector? I don't really know what I'm doing. I am not sure where the mistake is either.

Thank you for the reply :)
 
  • #4
Specter said:
How can I verify that the direction vectors are perpendicular to the normal vector?
Two vectors are perpendicular if their dot product is 0.
 
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  • #5
Mark44 said:
Two vectors are perpendicular if their dot product is 0.
Alright, so the dot product of the two direction vectors PQ and PR is not 0, meaning they are not perpendicular to the normal vector. How can I find out which direction vector is the vector causing the problems?
 
  • #6
Specter said:
Alright, so the dot product of the two direction vectors PQ and PR is not 0, meaning they are not perpendicular to the normal vector.
I think you misunderstood what I said. Find the dot product of each direction vector with the normal, not with each other. The two direction vectors don't have to be perpendicular, but each has to be perpendicular to the normal.
 
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  • #7
Mark44 said:
I think you misunderstood what I said. Find the dot product of each direction vector with the normal, not with each other. The two direction vectors don't have to be perpendicular, but each has to be perpendicular to the normal.

Yeah I did misunderstand.

PQ⋅n=0
PR⋅n=-6

So this tells me that PR does not work. Now what can I do to fix this?

Thanks!
 
  • #8
Specter said:
So this tells me that PR does not work. Now what can I do to fix this?
Check your work on getting PR. You have a mistake. Your point R is fine, but you made an error in calculating the vector PR.
 
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  • #9
Mark44 said:
Check your work on getting PR. You have a mistake. Your point R is fine, but you made an error in calculating the vector PR.

I checked my work for ##\vec PR## and got ##(-9,- \frac {19} 4, 5)##

Meaning that the vector equation would be ##(x,y,z)=(9,2,-5)+t(-9,-2,\frac {19} 6)+s(-9,-\frac {19} 4,5)## where s and t are any real numbers.

The parametric equations would be

##x=9-9t-9s
\\y=2-2t-\frac {19} 4s
\\ z=-5+\frac {19} 6t+5s##
 

Related to Find the scalar, vector, and parametric equations of a plane

1. What is a scalar equation of a plane and how is it different from a vector equation?

A scalar equation of a plane is an equation that represents a plane in terms of its distance from the origin and the orientation of the plane. It is different from a vector equation because it only uses scalar quantities (such as distance) rather than vector quantities (such as direction).

2. How do you find the normal vector of a plane?

To find the normal vector of a plane, you can use the cross product of two non-parallel vectors that lie on the plane. The resulting vector will be perpendicular to the plane and can be used as the normal vector in the plane's vector equation.

3. What are the steps for finding the parametric equations of a plane?

The steps for finding the parametric equations of a plane are as follows:

  1. Find the normal vector of the plane using the cross product of two non-parallel vectors on the plane.
  2. Choose a point on the plane to serve as the origin for the parametric equations.
  3. Select two direction vectors that are perpendicular to the normal vector and to each other.
  4. Write the parametric equations using the point from step 2 and the direction vectors from step 3.

4. Can you use a scalar equation to graph a plane in 3D space?

No, a scalar equation alone is not enough to graph a plane in 3D space. It only gives information about the distance and orientation of the plane, but not its position. To graph a plane, you will also need a point on the plane and at least one direction vector.

5. How can you tell if a point lies on a plane using its parametric equations?

To determine if a point lies on a plane using its parametric equations, you can plug in the coordinates of the point into the equations. If the resulting values satisfy the equations, then the point lies on the plane. If not, the point does not lie on the plane.

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