Determine the distance between the point $(2,-3,1)$ and the point of intersection of three planes

Question

Determine the distance between the point $(2,-3,1)$ and the point of intersection of the following system.

$3x-y+z=4$

$-x+2y+3z=7$

$x+3y+4z=12$

I'm not quite sure how to do this?

Answer 1

The hardest part is finding the point of intersection between the three planes. To do that, you can rewrite it as a matrix equation then apply Gaussian elimination, or you can just do it by hand.

Adding the $2$nd and $3$rd equations yields $$ 5y+7z=19. \tag{1}$$ Adding the $1$st equation and $3$ times the $2$nd equation gives $$ 5y+10z=25. \tag{2}$$ Subtracting $(1)$ from $(2)$ gives $$ 3z=6\iff z=2. $$ Substituting this into $(1)$ gives $$ 5y+7(2)=19\iff y=1. $$ Now, plugging in $y=1$ and $z=2$ into the first equation gives us $$ 3x-1+2=4 \iff x=1. $$ Thus, the point of intersection of the three planes is $(x,y,z)=(1,1,2).$ Now just find the distance between this point and $(2,-3,1)$.

Answer 2

Use Gaussian Elimination followed by the Euclidean Distance. If you don't know how to use either of these, read up on them from the links provided and from other sources(I recommend Lang's Linear Algebra as a introduction to the subject, but that is slightly off topic). The row echelon form of our system is:

$\left(\begin{array}{ccc|c} 1 & \frac{-1}{3} & \frac{1}{3} & \frac{4}{3} \\ 0 & 1 & 2 & 5 \\ 0 & 0 & 1 &2\end{array}\right)$

Solving this, we get $x = 1, y = 1, z = 2$. So we want the distance between $(1, 1, 2)$ and $(2, -3, 1)$ which by the distance formula is $$\sqrt{(1-2)^2 + (1+3)^2 + (2-1)^2} = \sqrt{1 + 16 + 1} = 3\sqrt{2}.$$