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For (a), I get z = -12x -25y - 63 For (b), I get that the constant are -4, -3, and 60 for x, y, and z respectively but, I don't know how to find the coefficients of t for x, y, and z and would appreciate it if someone could help me figure this out.
Basically, I get it all except the coefficients of t for part (b).
you might want to try part (a) again
My answer is correct. I checked.
did you type the question correctly....(-4,-3,60) is not a solution to the equation you gave above
Sorry, z != 2xy + 3y^2 + 2x + y + 4 z = x^2 + 2xy + 3y^2 + 2x + y + 4
do you know how to find the normal vector to the plane?
Is it -12 i - 25 j - 63k ?
and do you mean *a* normal vector?
no that is not a normal vector to the surface
So, I don't know. :(
find the coefficients infront of the x,y,z variables
Oh wait! Is the normal vector: 12 i + 25 j + k ?
(I did that based on what you just said.)
Oh so the coefficients of t are the coefficients of i, j, k.
I get it then :) Thanks!
don't forget to include your original point (-4,-3,60)