Show that the point $P$ of the line $f(t)=(at+\alpha, bt+\beta, ct+\gamma)$, $t \in \mathbb{R}$

Question

Show that the point $P$ of the line $f(t)=(at+\alpha, bt+\beta, ct+\gamma)$, $t \in \mathbb{R}$, closest to the origin is that $P=f(t_{0})$ with $t_{0}=-\frac{a\alpha+b\beta+c\gamma}{a^{2}+b^{2}+c^{2}}$.

Answer 1

Let $P = f(t)$ be the point on the given line closest to the origin.

The distance $D$ between $P$ and the origin is given by

$$D^2 = (at+\alpha)^2 + (bt+\beta)^2 + (ct+\gamma)^2$$

Or, $$2D \frac{dD}{dt} = 2a(at+\alpha) + 2b(bt+\beta) + 2c(ct+\gamma)$$

To minimize $D$, let us set $$\frac{dD}{dt} = 0$$

Or, $$a(at+\alpha) + b(bt+\beta) + c(ct+\gamma) = 0$$

Or, $$t = -\frac{a\alpha+b\beta+c\gamma}{a^{2}+b^{2}+c^{2}}$$

Answer 2

It suffices to show the vector from the origin to $P$ is orthogonal to the given line. This vector is given by $$\frac{1}{a^2+b^2+c^2}(-a b B - a c C + A b^2 + A c^2, a^2 B - a A b - b c C + B c^2, a^2 C - a A c + b^2 C - b B c).$$ Taking the dot product of this with the direction numbers $(a,b,c)$ for the line yields $0$ as desired.