Question

Find the equation of the tangent plane at $(1,1,1)$ to $x^2+y^2+z^2-3=0$.

Answer 1

Find the required equation:

Given: the equation of the sphere $x^2+y^2+z^2-3=0$, point $(1,1,1)$.

We know that the equation of tangent plane to the sphere $x^2+y^2+z^2+ux+vy+wz+d=0$ at $(x_1,y_1,z_1)$ is,

$x{x}_1+y{y}_1+z{z}_1+u(x+x_1)+v(y+y_1)+w(z+z_1)+d=0$

So the equation of tangent plane at $(1,1,1)$ to $x^2+y^2+z^2-3=0$ is,

$$\begin{gathered} x\left(1\right)+y\left(1\right)+z\left(1\right)–3=0 \\ \Rightarrow x+y+z=3 \end{gathered}$$

Therefore, the required equation of the tangent plane is $x+y+z=3$.