Need help finding the minimal distance between two points

Question

Hello I've been doing an exercise but I don't know how to find the answer. I need your help, please.

Question

Let Q(1,1,4) be a point on the space and P(x,y,z) any point on the line Delta : (x,y,z) = (3,0,4) + t (2,1,-1), where t belongs to R. For which value of t is the distance between point Q and point P minimal?

What I've done

I determined the coordinates of P

I determined the length of the director vector d and the components of PQ

I determined the expression of PQ×d

I found the expression of the distance

Answer 1

The square of the distance between $P$ and $Q$ is

$((3+2t)-1)^2+((t)-1)^2+((4-t)-4)^2=6t^2+6t+5=\frac32(2t+1)^2+\frac72.$

This is minimized when $2t+1=0$ or $t=-\frac12$.

Answer 2

To stick to your geometric approach using orthogonality you may proceed as follows:

• Let $P_t = P + td$ with $d = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$.
• $||Q-P_t||$ is minimal iff $(Q-P_t) \perp d$.

Hence,

$$(Q-P_t) \perp d \Leftrightarrow (Q-P-td)\cdot d = 0$$ $$\Leftrightarrow (Q-P)\cdot d =t||d||^2 \Leftrightarrow \frac{(Q-P)\cdot d}{||d||^2} =t $$

Plugging in the given values gives

$$t = \frac{\begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix}\cdot\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}}{2+1+1} = -\frac 12$$