Surface tangent plane

Question

Determine a plane that is tangent to the surface $x ^{2}$ + $3y ^ 2$ + $2z ^ 2$ = $\frac{11}{6}$ and parallel to the plane x + y + z = 10

$$\nabla f (x, y, z) = (2x, 6y, 4z)$$

Quotation vector $X + Y + Z = 10 \implies (1,1,1)$

$$(2x, 6y, 4z) = K (1,1,1)$$

for $K = 1$

• $2x = 1 x = \frac{1}{2}$

• $6y = 1 y = \frac{1}{6}$

• $4z = 1 z = \frac{1}{4}$

Putting the plane equation together: $(1,1,1) [(x - 1/2) + (y-1/6) + (z-1/4)]$

My plan equation gave: $X + Y + Z = \frac{-11}{12}$

But the answer of the book is $X + Y + Z = \frac{-11}{6}$or X + Y + Z = $\frac{11}{6}$

Answer 1

You cannot impose that $K=1$. You're after a point $(x,y,z)$ of the surface $x^2+3y^2+2z^2=\frac{11}6$ which is of the form $\left(\frac K2,\frac K6,\frac K4\right)$. So, you solve the equation$$\left(\frac K2\right)^2+3\left(\frac K6\right)^2+2\left(\frac K4\right)^2=\frac{11}6.$$You will get $K=\pm2$. Can you take it from here?

Answer 2

I think there is an error for $x$ and $y$, I obtain $K=\pm 2$ and

• $x=\pm1$
• $y=\pm \frac13$
• $z=\pm \frac12$