# Dot product: normal and tangent

1. Mar 17, 2009

### sci-doo

1. The problem statement, all variables and given/known data
Tangent plane goes through point P=(a,b,f(a,b)). Any point on the plane is then
Q=(x,y,z)=(x,y,f(a,b)+fx(a,b)(x-a)+fy(a,b)(y-b)) (fx and fy are partial derivatives)
and the vector $$\overline{PQ}$$ is on tangent plane.
Calculate dot product n.$$\overline{PQ}$$ and show that the normal vector is perpendicular to the tangent plane.

2. Relevant equations

n=fx(a,b)$$\hat{i}$$ + fy(a,b)$$\hat{j}$$ - $$\hat{k}$$

3. The attempt at a solution
Ok, I should get the dot product = 0.

I don't know how to do that, because I don't know how to get the vector $$\overline{PQ}$$ into a right form. So far I've tried

$$\overline{PQ}$$ = $$\overline{Q}$$ - $$\overline{P}$$
= xi + yj +f(x,y)k - ai - bj - f(a,b)k
= (x-a)i + (y-b)j + (f(x,y)-f(a,b))k

How am I supposed to get zero dot product from THAT and n? :yuck:
n has partial derivatives, can I write them some other way? I'm stuck.

2. Mar 17, 2009

### Staff: Mentor

So start with actually doing the dot product of PQ and n. What do you get?

3. Mar 17, 2009

### sci-doo

I get

n dot PQ = fx(a,b)(x-a)+fy(a,b)(y-b)-f(x,y)+f(a,b)

I still have those partial derivatives, and I don't know what to do with them.

4. Mar 17, 2009

### Staff: Mentor

That's not what I get. You have an error in what you got for PQ.

5. Mar 17, 2009

### sci-doo

To get PQ I think I should start with Q-P. But it is scalar and I don't see how it turns into a vector.

Q-P= x,y,f(a,b) + fx(a,b)(x-a) + fy(a,b)(y-b) - a,b,f(a,b)

6. Mar 17, 2009

### Staff: Mentor

Q is a vector and P is a vector, and Q - P is a vector, and is equal to PQ. When you add or subtract vectors, you do so component-wise.

7. Mar 17, 2009

### sci-doo

Q and P are given as scalars. P=(a,b,f(a,b)) is, I think, relatively easy to vectorize
ai + bj + f(a,b)k
Q is a bit ugly
xi + yj + f(a,b)k + fx(a,b)xi - fx(a,b)ai + fy(a,b)yj - fy(a,b)bj
and so Q - P is
xi + yj + fx(a,b)xi - fx(a,b)ai + fy(a,b)yj - fy(a,b)bj - ai - bj
and as k is gone, can be put as i and j
(x + fx(a,b)x - fx(a,b)a -a)i + (y + fy(a,b)y - fy(a,b)b -b)j

Taking dot product with n :

(x + fx(a,b)x - fx(a,b)a -a)*(fx(a,b)) + (y + fy(a,b)y - fy(a,b)b -b)*(fy(a,b)) = ..

The product doesn't cancel out, it is not zero.

8. Mar 17, 2009

### Tom Mattson

Staff Emeritus
I don't know how you're getting that $\vec{PQ}$, but the dot product with the normal vector is obviously zero.

???

How do you get terms such as $f_x(a,b)x\hat{i}$?

$f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)$ is the z coordinate of Q.

9. Mar 17, 2009

### Staff: Mentor

The expression above is incorrect. Your x and y components are correct, but all the rest is the multiplier for the unit vector k.

10. Mar 17, 2009

### sci-doo

Thank you for correcting my somewhat unusual (and wrong!) approach..

I didn't see the z-part of Q as it is.

Dot product is (of course):
$$\bar{PQ}$$.n = fx(a,b)(x-a) + fy(a,b)(y-b) - fx(a,b)(x-a) - fy(a,b)(y-b) = 0

Once again, thank you for your help.