Normal vectors and tangent planes

Question

Could you check my work please? Let me know if it's right or wrong.

We have the level surface $$f(x, y, z) = xyz -6$$

The normal vector is equal to the gradient, so at the point $(a, b, c)$ $$\nabla f = \langle f_x, f_y, f_z \rangle$$

$$\:\:\:\:\:\:\:=(yz, xz, xy)$$

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=bc \: \vec{i} + ac \: \vec{j} + ab \: \vec{k}$$

The tangent plane is

$$ F_x(x_0, y_0, z_0)(x - x_0) + F_y(x_0, y_0, z_0)(y - y_0) + F_z(x_0, y_0, z_0)(z - z_0) = 0 $$

At the point $(a, b, c)$

$$ bc (x - a) + ac(y - b) + ab(z - c) = 0 $$

$$ bcx+acy+abz = 3abc $$

score 1 · Accepted Answer · answered Oct 24 '14 at 02:07

1

Your gradient vector is wrong. You have to directly input the values of the vector $(a,b,c)$ as $x,y,z$, respectively. So the proper value of $\nabla f = (yz,xz,xy)$ should be $(bc,ac,ab)$, besides that, everything seems ok.

answered Oct 24 '14 at 02:07

Rono

1,039

I tried to fix it, is that what you meant? – Jack Oct 24 '14 at 02:11
btw I haven't done this in class yet, I tried to teach myself online – Jack Oct 24 '14 at 02:13
@Jack, yep, actually if you work around the final value of the plane, you get a more interesting equation: $bcx+acy+abz = 3abc$ – Rono Oct 24 '14 at 02:13
Oh gosh you're right, thanks :) – Jack Oct 24 '14 at 02:16
Hi @Jack please can you help with https://math.stackexchange.com/q/4396103/585488 – linker Mar 05 '22 at 15:24

score 1 · Answer 2 · answered Oct 24 '14 at 02:15

1

Right now, I believe your solution is correct. Every time you do this kind of question, just do the gradient first and then plug it into the formula of tangent plane at a point, then you will get the correct answer.

answered Oct 24 '14 at 02:15

BabyMiu

15

Normal vectors and tangent planes

2 Answers2