Find the distance from the point $P(3,-1,4)$ and the line whose parametric equations are: $x=-2 + 3t$, $y=-2t$, and $z=1+4t$
I'm not completely sure how to solve this so I first gave the parameter $t$ some initial values:
$t=0 = Q(-2,0,1)$
$t=1 = R(1,-2,5)$
$\overrightarrow {QR} = \langle 1+2, -2-0,5-1\rangle = \langle 3,-2,4 \rangle$
$\overrightarrow {QP} = \langle 3+2, -10-, 4-1\rangle = \langle 5, -1,3 \rangle$
$\overrightarrow {QR} \times \overrightarrow {QP} = -2i+11j+7k$
$d=\frac{|\overrightarrow {QR} \times \overrightarrow {QP}|}{|\overrightarrow {QR}|} = \frac{\sqrt{166}}{\sqrt{21}}$
Is this correct?
Best Answer
Let $S(-2+3t,-2t,1+4t)$ be a point on the line such that $PS$ is perpendicular to the line.
Then, we have $\vec{PS}=(-2+3t-3,-2t-(-1),1+4t-4)=(3t-5,-2t+1,4t-3)$. This is perpendicular to the vector $(3,-2,4)$, we have $$(3t-5)\cdot 3+(-2t+1)\cdot(-2)+(4t-3)\cdot 4=0\iff t=1.$$
Thus, the minimum distance is the distance between $S(1,-2,5)$ and $P(3,-1,4)$, i.e. $$\sqrt{(1-3)^2+(-2+1)^2+(5-4)^2}=\color{red}{\sqrt{6}}.$$