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## s3a 4 years ago (a) What is the equation of the tangent plane to the surface z = 2xy + 3y^2 + 2x + y + 4 at the point P(-4,-3,60)? (b) What are the parametric equations of the normal line at this point?

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1. s3a

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.

2. s3a

Basically, I get it all except the coefficients of t for part (b).

3. Zarkon

you might want to try part (a) again

4. s3a

My answer is correct. I checked.

5. Zarkon

did you type the question correctly....(-4,-3,60) is not a solution to the equation you gave above

6. s3a

Sorry, z != 2xy + 3y^2 + 2x + y + 4 z = x^2 + 2xy + 3y^2 + 2x + y + 4

7. s3a

(I'm sleepy.)

8. Zarkon

do you know how to find the normal vector to the plane?

9. s3a

Is it -12 i - 25 j - 63k ?

10. s3a

and do you mean *a* normal vector?

11. Zarkon

no that is not a normal vector to the surface

12. s3a

So, I don't know. :(

13. Zarkon

$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ is a plane with normal vector $$\vec{n}=<a,b,c>$$

14. Zarkon

find the coefficients infront of the x,y,z variables

15. s3a

Oh wait! Is the normal vector: 12 i + 25 j + k ?

16. s3a

(I did that based on what you just said.)

17. Zarkon

yes

18. s3a

Oh so the coefficients of t are the coefficients of i, j, k.

19. s3a

I get it then :) Thanks!

20. Zarkon

don't forget to include your original point (-4,-3,60)

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