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

[Physics345]

Homework Statement

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

Homework Equations

Ax+By+Cz+D=0
(x,y,z)=(x0,y0,z0)+s(a1,a2,a3)+t(b1,b2,b3)
x=x0+sa1+tb1
y=y0+sa2+tb2
z=z0+sa3+tb3

The Attempt at a Solution

Scalar Equation:
3x-4y+6z+D=0
3(9)-4(2)+6(-5)+D=0
D=11
3x- 4y +6z+11=0
Vector Equation:
4y=3x+6z+11
y=3/4 x+6/4 z+11
y=3/4 x+3/2 z+11
P(9,2,-5)
Let x=0 and z=0
y=11
Q(0,11,0)
let x=2 and z=1
y=3/4(2)+3/2(1)+11
y=14
R(2,14,1)
(PQ) ⃗ =Q-P
(PQ) ⃗ =(0,11,0)-(9,2,-5)=(-9,9,5)
(PR) ⃗ =R-P
(PR) ⃗ =(2,14,1)-(9,2,-5)=(-7,12,6)
(x,y,z)=(9,2,-5)+s(-9,9,5)+t(-7,12,6)
Parametric Equations:
x=9-9s-7t
y=2+9s+12t
z=-5+5s+6t
Did I do this correctly? And is there anyway to confirm my answers through math? Thanks in advance =)

 

[SammyS]

Homework Statement

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

Homework Equations

Ax+By+Cz+D=0
(x,y,z)=(x0,y0,z0)+s(a1,a2,a3)+t(b1,b2,b3)
x=x0+sa1+tb1
y=y0+sa2+tb2
z=z0+sa3+tb3

The Attempt at a Solution

Scalar Equation: is OK.

Vector Equation:
4y=3x+6z+11
y=3/4 x+6/4 z+11

You did not divide 11 by 4 .

y=3/4 x+3/2 z+11
P(9,2,-5)
Let x=0 and z=0
y=11
Q(0,11,0)

That point is not in the plane.

let x=2 and z=1
y=3/4(2)+3/2(1)+11
y=14
R(2,14,1)
(PQ) ⃗ =Q-P
(PQ) ⃗ =(0,11,0)-(9,2,-5)=(-9,9,5)
(PR) ⃗ =R-P
(PR) ⃗ =(2,14,1)-(9,2,-5)=(-7,12,6)
(x,y,z)=(9,2,-5)+s(-9,9,5)+t(-7,12,6)
Parametric Equations:
x=9-9s-7t
y=2+9s+12t
z=-5+5s+6t
Did I do this correctly? And is there anyway to confirm my answers through math? Thanks in advance =)

The parametric version is also incorrect.

 

[Physics345]

Scalar Equation: is OK.You did not divide 11 by 4 .That point is not in the plane.
The parametric version is also incorrect.

Which point are you referring to from PQR? Is it because of the division mistake?

 

[SammyS]

Which point are you referring to from PQR? Is it because of the division mistake?

I am referring to the point Q at (0, 11, 0) .

That does appear to me to be the result of your division mistake.

 

[Physics345]

Okay according, to your comments I made some adjustments, how are you able to tell by the way?

y=3/4 x+3/2 z+11/4
P(9,2,-5)
Let x=0 and z=0
y=11/4
Q(0,11/4,0)
let x=2 and z=1
y=3/4(2)+3/2(1)+11/4
y=23/4
R(2,23/4,1)
(PQ) ⃗=Q-P
(PQ) ⃗=(0,11/4,0)-(9,2,-5)=(-9, 3/4,5)
(PR) ⃗=R-P
(PR) ⃗=(2,23/4,1)-(9,2,-5)=(-7,15/4,6)
(x,y,z)=(9,2,-5)+s(-9, 3/4,5) +t(-7,15/4,6)
Parametric Equations:
x=9-9s-7t
y=2+3/4 s+15/4 t
z=-5+5s+6t

 

[SammyS]

Okay according, to your comments I made some adjustments, how are you able to tell by the way?

What is the thing that you are asking about? I made four different statements in my post #2 of this thread.

 

[Physics345]

What is the thing that you are asking about? I made four different statements in my post #2 of this thread.

That the points were wrong. Also is it right now?

 

[SammyS]

Okay according, to your comments I made some adjustments, how are you able to tell by the way?
...

(x,y,z)=(9,2,-5)+s(-9, 3/4,5) +t(-7,15/4,6)

Parametric Equations:
x=9-9s-7t
y=2+3/4 s+15/4 t
z=-5+5s+6t

That the points were wrong. Also is it right now?

Your results are now correct.

As far as checking your results:
I knew that your scalar equation for the plane was correct, by verifying the steps you took.

Using the scalar equation, it's easy to check that a given point is in the plane. Plug-in x, y, and z .

 

[Physics345]

P(9,2,-5)
3x- 4y +6z+11=0
3(9)-4(2)+6(-5)+11=0
=0
How did it verify? because it equals zero?

 

[SammyS]

P(9,2,-5)
3x- 4y +6z+11=0
3(9)-4(2)+6(-5)+11=0
=0
How did it verify? because it equals zero?

Yes. Plugging into the equation gave a "true" statement.

 

[Physics345]

Yes. Plugging into the equation gave a "true" statement.

Thank you, I understand now. Your help was very much appreciated.