# 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!

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Could you write in English? – user17762 Jun 25 '12 at 2:21
Consider the line L defined by the following parametric equations x=3+2t y=4+t z=5−6t I want to find the point Q that lies on L that is closest to (4,1,7). Note: Really, I don't remember the formulas, I need help! – MGN Jun 25 '12 at 2:25
@marvis There has been strong consensus on meta in the past that people generally prefer coherent questions in foreign languages, which can be intelligently translated to English (as this one was) to incoherent, unintelligible questions in English. – MJD Jun 25 '12 at 18:00

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.

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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 $(4,1,7)$.

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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.

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 $(97/41,151/41,283/41)$.