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So I am given the points $A=(1,3,-2)$ , $P_1=(2,0,-1)$ and $P_2=(4,-2,-1)$. I am asked to find the point $P$ on the line through $P_1$ and $P_2$ that is closest to $A$.

I would appreciate some guidance as I am not too sure as to how to actually approach this. Thank you very much.

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    Hint: $AP\perp P_1P_2$. – 2017-01-26
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    Yep that's the only hint I needed haha. Thanks. I got it now. – 2017-01-26

2 Answers 2

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A generic point on the $P_1 P_2$ line has the following form: $$ P_\lambda = \lambda P_1 + (1-\lambda)P_2 = (4-2\lambda,2\lambda-2,-1)$$ and the squared distance from $A$ is given by $$ \|P_\lambda-A\|^2 = (2\lambda-3)^2+(2\lambda-5)^2 +1^2 $$ so the distance is minimal when $$ 8\lambda^2-32\lambda+35 $$ is minimal, i.e. at $\lambda=2$. In such a case $P=P_\lambda=\color{red}{(0,2,-1)}$ and the distance is $\color{red}{\sqrt{3}}$.

$AP\perp P_1 P_2$ is a consequence of the minimality of $AP_\lambda$.

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    I did it another way using projections but I got the same answer so thanks for confirming that =] . – 2017-01-27
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I would say:

  1. Find the vector created by the two points
  2. Find the equations (2) of a straight line that is generated by this vector and passes through P1 (and necessarily P2)
  3. Find the equation (1) of the plane that is perpendicular to that vector and passes through A
  4. Intersect the line and the plane by solving the system of 3 equations, you will find your point (3 unknowns).

Good luck!