Question

How to find the vector equation of the line by the point tangent to the plane

Answer 1

Answer:Well, this is obviously your online homework, so I'll avoid giving you the whole thing. To get the coefficient of x to be 1, use the level surface

F(x, y, z) = x - y² - 2z² = 28.

Then ∇F = <1, -2y, -4z>. It should be exceedingly easy to finish from here.

**** Update ****

Not sure how you'd get the wrong answer. At the point (5, -1, -4)

∇F = <1, 2, 16>

The plane containing (a, b, c) with normal vector <A, B, C> has equation

A(x - a) + B(y - b) + C(z - c) = 0.

Plug in A = 1, B = 2, C = 16 and (a, b, c) = (5, -1, -4).

The normal line is

<5, -1, -4> + t<1, 2, 16>.

There is nothing to this.

Step-by-step explanation: