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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.