- #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.