Find the equation of the two tangent planes to the sphere $x^2+y^2+z^2-2y-6z+5=0$ which are parallel to the plane $2x+2y-z=0$
My Attempt
We need to find a point which is shortest distance from the plane to the sphere. Let it be $(a,b,c).$ Then Equation of the tangent plane at $(a,b,c)$ is given by the formula
$$2a(x-a)+2(b-1)(y-b)+2(c-3)(z-c)=0$$
Finding the diametrically opposite point of $(a,b,c)$. I can find the equation of another tangent parallel to the plane.
I don't know how to find the shortest distance from the plane to the sphere.
Is there any short method to find the tangents?
Now, let $\vec{r}$ be the position vector of any point on the given plane. From the figure it's easy to observe that NP is perpendicular to ON and therefore: $\vec{NP}\cdot\vec{ON}=0$$$(r-d\hat{n})\cdot d\hat{n}=0$$Simplifying the above equation, you get:$$\vec{r}\cdot\hat{n}=d$$Which is known as the normal form equation of plane.(Note that unit vector of normal is required).
Hence, if $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$, normal vector is $\vec{n}=a\hat{i}+b\hat{j}+c\hat{k}$, and the distance of plane from origin is $d$, then first find unit vector along normal which comes out to be $\frac{\vec{n}}{|n|}$
Now equation of plane is $$r\cdot\frac{\vec{n}}{|n|}=d$$which can be written as$$ax+by+cz=d\cdot|n|$$This is the Cartesian form of the plane.
Best Answer
Instead complete the square
$$x^2+(y-1)^2+(z-3)^2 = 5$$
then take the gradient
$$\langle x, y-1, z-3 \rangle = \lambda\langle 2, 2, -1\rangle \implies x = y-1 = \frac{3-z}{2}$$
which means
$$x^2 + x^2 + 4x^2 = 6x^2 = 5 \implies x = \pm \sqrt{\frac{5}{6}}$$
This gives you your two points once you plug in for $y$ and $z$.