Question

At what point on the paraboloid
y = x2 + z2
is the tangent plane parallel to the plane
7x + 2y + 3z = 2?
(If an answer does not exist, enter DNE.)

Answer 1

normal vector to f(x,y,z)=0 is

so, for the paraboloid,
<b>n</b> = f_x<b>i</b> + f_y<b>j</b> + f_z<b>k</b>
= 2x<b>i</b> - <b>j</b> + 2z<b>k</b>

for the plane,
<b>n</b> = 7<b>i</b> + 2<b>j</b> + 3<b>k</b>

Now manipulate the directions and magnitudes so the two normals are the same.

Answer 2

In other words,

divide the plane's normal vector so that the y components match (INCLUDING SIGN), then solve for the two remaining variables.

For the example above this would mean;

2xi-j+2zk=7i+2j+3k
becomes
2xi-j+2xk=(-7/2)i-j+(-3/2)k

Now solve for x and z like so:
2x=(-7/2) => x=(-7/2)/2 = -7/4

2z=(-3/2) => z=(-3/2)/2 = -3/4

Now for y, just use the original equation (y=x^2+z^2), since we now know x and z, simply plug them in to get y like so:

y= (-7/4)^2+(-3/4)^2 = 29/8

So your coordinates are:

(-7/4, 29/8, -3/4)

Happy trails.