Point on a line with the least distance from another point in $\mathbb{R}^3$ Consider the line $L$ defined by the following parametric equations
$$x= 3+2t$$
$$y= 4+t$$
$$z=5-6t$$
Find the point $Q$ on $L$ that is closest to $(4,1,7)$.
Note: I do not really remember the formulas, I need help!
 A: The plane containing this closest point will have normal vector
$$n = 2i + j - 6k. $$
Since the point $(4,1,7)$ is in the plane and the plane's equation is
$$ 2x + y - z = 2\cdot4 + 1 - 7 = 1.$$
The closest point will lie where the point and line intersect. Find this, then the distance to the point $(4,1,7)$.
A: The distance between $(x,y,z)$ and $(4,1,7)$ is $\sqrt{(x-4)^2+(y-1)^2+(z-7)^2}$.  That follows from the Pythagorean theorem.  This is the same as
$$\sqrt{(2+2t-4)^2+(4+t-1)^2+(5-6t-7)^2}.$$
That simplifies to
$$
\sqrt{(2t-2)^2+(3+t)^2+(-2-6t)^2}.
$$
The value of $t$ that minimizes the distance is the same as the value of $t$ that minimizes the square of the distance, i.e. $(2t-2)^2+(3+t)^2+(-2-6t)^2$.  If you multiply that out, you get $(\bullet t^2) + (\bullet t) + (\bullet)$.  Find the three numbers and then complete the square.
A: Thanks to Michael Hardy, in (poor) spanish:
La distancia entre $(x,y,z)$ y $(4,1,7)$ es $\sqrt{(x-4)^2+(y-1)^2+(z-7)^2}$.  Este sigue de la teorema de Pythagorean.  Esto es lo mismo que
$$\sqrt{(2+2t-4)^2+(4+t-1)^2+(5-6t-7)^2}.$$
Simplifica a
$$
\sqrt{(2t-2)^2+(3+t)^2+(-2-6t)^2}.
$$
El valor de $t$ que minimiza la distancia es lo mismo que el valor de $t$ que minimiza el el cuadrado de la distancia: $(2t-2)^2+(3+t)^2+(-2-6t)^2$.  Si multiplica eso, consique $(\bullet t^2) + (\bullet t) + (\bullet)$.  Encuentra estos numeros y completa el cuadrado.
Tambien, disculpa mi traduccion mala.
A: Direction cosines of given line are $(2,1,-6)$. If $P=(4,1,7)$ and $Q=(3+2t,4+t,5-6t)$ be the closest point on the line to P, then line joining these two points must be normal to direction cosines of given line. Hence, dot product of vectors $PQ=(2t-1,3+t,-6t-2)$ and $(2,1,-6)$ would be $0 \implies$ $4t-2+3+t+36t+12=0 \implies 41t=-13 \implies t=-13/41$ for $Q$.Thus, the point is  $(97/41,151/41,283/41)$.
