Determine the distance from the point $(1, 3, 2)$ to the intersection line of the planes $x + y + 2z = 1$ and $x – z = 2$.
That's what i did to solve the question:
First i determined the intersection of planes $x + y + 2 z = 1$ and $x – z = 2$.
Then i got that:
$x = -t$ and
$y = -1-3t$ and
$z = 2t$.
Now we can calculate the distance from point $A = (1,3,2)$ and line $(x,y,z) = (0,-1,0)+ t (-1,-3,2)$.
Then i used the formula $d=\frac{v\;\mathrm X\;pA}{|v|}$
I know that $pA = (1,4,2)$ but how about $|v \;\mathrm X\; pA|$? It's $(-14,4,-1)$ right?
So the distance will be $\sqrt \frac{213}{14}$ right?
Best Answer
Your method is correct, though it is noted in the comment you may have a slight algebra error. Here is another method, which requires some calculus but does not make use of a predetermined formula.
You have correctly identified that you wish to compute the distance from the point $p=(1,3,2)$ to the line $$ L(t)=(-t,-1-3\,t,2\,t) $$ The distance from $p$ to the point $L(t)$ is given by the function $$ \DeclareMathOperator{dist}{dist}\dist(t)=\sqrt{(1+t)^2+(4+3\,t)^2+(2-2\,t)^2} $$ so we need only minimize $\dist(t)$. But, since $\dist(t)$ and $D(t)=\dist(t)^2$ share critical points, we can instead minimize $D(t)$ which is easier.
To do so, note that \begin{align*} D^\prime(t) &= 2\,(1+t)+2\,(4+3\,t)(3)+2\,(2-2\,t)(-2) \\ &= 28\,t+18 \\ D^{\prime\prime}(t) &= 28 \end{align*} This implies that the (global) maximum of $D(t)$ and hence $\dist(t)$ occurs at $t=-9/14$. Our answer is then $$ \dist\left(-\frac{9}{14}\right)=\sqrt{\frac{213}{14}} $$