How do I find the intercept of X , given three points $P(x,y,z) , Q(x,y,z) , R(x,y,z)$ in a plane?

Question

I have three points in a plane, i need to find where it intercepts in the x axis , how do I find the coordinate that does that? I have to figure out the plane equation first?

Answer 1

No, you don't have to figure out the plane equation.

Let $X(x_0,0,0)$ be the intersection point:

We have:

$$\begin{vmatrix}x_0&x_P&x_Q&x_R\\ 0&y_P&y_Q&y_R\\ 0&z_P&z_Q&z_R\\ 1&1&1&1 \end{vmatrix}=0$$

Why that ? Because $\tfrac16$ of this determinant is equal to the volume of tetrahedron $PQRX$ (see here) and we want this volume to be $0$.

It remains to expand this determinant with respect to the first column and solve for $x_0$:

$$x_0=\dfrac{\begin{vmatrix}x_P&x_Q&x_R\\ y_P&y_Q&y_R\\ z_P&z_Q&z_R\\ \end{vmatrix}}{\begin{vmatrix} y_P&y_Q&y_R\\ z_P&z_Q&z_R\\ 1&1&1 \end{vmatrix}}$$

with the condition that the denominator is not zero, of course.