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

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  • #1
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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.
 

Answers and Replies

  • #2
35,062
6,794

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.
 
  • #3
120
8
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
35,062
6,794
How can I verify that the direction vectors are perpendicular to the normal vector?
Two vectors are perpendicular if their dot product is 0.
 
  • #5
120
8
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
35,062
6,794
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.
 
  • #7
120
8
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
35,062
6,794
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.
 
  • #9
120
8
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##
 

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