L is the Line with parametric equation:


Find the shortest distance d from point =(3,1,1) to L and point Q on L that is closest to the point.

for the distance I get: (((t+5)^2)+((t+5)^2)+((t+3)^2))^(1/2)

I'm not sure if this is right for the distance.

  • $\begingroup$ How did you get this answer? Please write out your proof. $\endgroup$ – Brevan Ellefsen Mar 18 '19 at 3:55
  • $\begingroup$ You have written down the square of the distance from the given point to an arbitrary point on the line (except that $3-2\ne5$), but what's wanted is the distance to the closest point on the line. $\endgroup$ – Gerry Myerson Mar 18 '19 at 3:56

There's probably a much easier way to solve this, but we have the line $$L = \langle(2-t), (-4+t), (-2-t)\rangle$$ and the point $P=(3,1,1)$. The distance at any time $t$ between said line and point is $$D=\sqrt{(t+1)^2+(5-t)^2+(3+t)^2}$$ Minimizing the distance is the same as minimizing the square of the distance (this can be simply explained: distance is always positive). $$D_2=(t+1)^2+(5-t)^2+(3+t)^2$$ $$(D_2)'=2(t+1)-2(5-t)+2(3+t)$$ The distance is minimized when the derivative is $0$. $$0=2t+2-10+2t+6+2t$$ $$0=6t-2$$ $$t=\frac13$$ So, at time $t=\frac13$, $\,L(\frac13)=\langle \frac53,-\frac{11}3,-\frac73\rangle$ And the distance $D$ is (once you plug and chug) $\frac{53}9$


The line is parallel to $u=\langle-1,1,-1\rangle$.

A point on the line is $(2,-4,-2)$.

The vector from that point to $(3,1,1)$ is $v=\langle1,5,3\rangle$.

Picture these two vectors $u$ and $v$. If you orthogonally project $v$ onto the span of $u$, you get to the nearest point that you are hunting (relative to the base point $(2,-4,-2)$). To project, $$\operatorname{proj}_{u}v=\frac{u\cdot v}{u\cdot u}u=\frac{1}{3}\langle-1,1,-1\rangle$$ Now add this to $(2,-4,-2)$ and you have the closest point: $$\left(\frac53,-\frac{11}{3},-\frac73\right)$$


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