Find the scalar equation of the plane that passes through the line of intersection of two planes: $x+y+z-4 = 0$ and $y+z-2=0$ and the plane is 3 units from the point (5,-3,7).
I am able to form a plane that goes through the line of intersection of two planes by using a linear combination:
$$x+y+z-4 + k(y+z-2)=0$$
where $k$ is real number.
How do I find the value of k so that the plane is 3 units from the point?
Best Answer
You’ve got a good start, although you really need to use a more general linear combination of the two planes instead: $$\lambda(x+y+z-4)+\mu(y+z-2)=0. \tag 1$$ The problem with your family of planes is that they exclude $y+z-2=0$ itself. That would leave you stuck with no solution if that plane happened to be the correct one. You can easily verify that the distance from $(5,-3,7)$ to this plane is not $3$, though, so in this case you can proceed with your original formulation.
Now you can simply use the formula for the distance of a point $P=(x_p,y_p,z_p)$ from the plane $ax+by+cz+d=0$: $${|ax_p+by_p+cz_p+d| \over \sqrt{a^2+b^2+c^2}}.$$ Plug in the coordinates of the point from the problem and use (1) for the plane, and set this equal to the required distance from the point. Squaring both sides and rearranging will give you a quadratic equation in $k$ to solve.