Points and a tangent plane

Question

The task is: "Find the point(s) on $x^2+y^2+z^2 = 8$, where the tangent plane is parallel to the plane $x-y+2z=0$"

What I did is: $f(x,y,z) = x^2+y^2+z^2-8$ grad$f = (2x,2y,2z)$

normal vector of $x-y+2z=0$ is $n=(1,-1,2)$

grad$f \parallel n \iff 2x=1, 2y=-1, 2z=2$

Therefore wanted point is $(\frac{1}{2},-\frac{1}{2};1)$ Is this ok?

I don't think that's quite right, the point you want needs to be on the sphere $x^2+y^2+z^2 = 8$ but $\frac{1}{2}^2+(\frac{-1}{2})^2+1^2 \not = 8$. One idea you can play around with is that the unit normal to a sphere at a point $(x,y,z)$ on the sphere is just the normal vector associated with $(x,y,z)$, so if you want to see what point on the sphere has normal $<1,-1,2>$ just extend the line from this vector until it intersects the sphere and you'll have your two points. (because you'll intersect the sphere twice) — user413766, Jun 19 '18 at 13:02

score 0 · Accepted Answer · edited Jun 12 '20 at 10:38

You're on the right track.

grad$f = (2x,2y,2z)$

normal vector of $x-y+2z=0$ is $n=(1,-1,2)$

grad$f \parallel n \iff 2x=1, 2y=-1, 2z=2$

Note that for the gradient of $f$ to be parallel to the normal vector, the vectors don't have to be equal but scalar multiples; so you get: $$\mbox{grad}\,f \parallel n \iff (2x,2y,2z)=k\cdot(1,-1,2)$$ Substitution of $x$, $y$ and $z$ as a function of $k$ into the equation of the sphere gives you a quadratic equation in $k$; the solutions correspond to points on the sphere with the desired gradient.

Points and a tangent plane

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