Equation of line passing through two given lines (3D) A line from the origin meets the line $x-2\over 1$ = $y-1\over-2$ = $z+1\over1$ And $(x-\frac83)\over2$ = $y+3\over-1$ = $z-1\over1$ at P and Q respectively. If the length $PQ = d$ , then $(d)^2$ is
My attempt :
I assumed general points on the given lines, thus introducing two variables, say a and b. Then I found out the direction ratio of the required line in terms of a and b, (by subtracting the pairs of each of the three points). The required line would be a line with the already found out DR and passing through the assumed point on line 1(in terms of a). I wrote the expression for a general point on the required line by introducing a third variable , say c , and then equating each of the variables to zero as the line passes through origin. 3 equation and three variables.
But solving this is very lengthy. After which we would have to apply distance formula as well. I am looking for a smarter and shorter approach,
Thank you
This is also my first question on this site
 A: Let point $P$ be on the first line, then
$P = P_0 + t V $
And let point $Q$ be on the second line, then
$Q = Q_0 + s W $
Since we want the line extending from $P$ to $Q$ to pass through the origin, then it follows that $P$ is on the plane that passes through the origin and contains $Q(s)$.  Similarly, $Q$ is on the plane that passes through the origin and contains the line $P(t)$.
There is a single plane that passes through the origin and contains the first line.  The equation of this plane is
$ ( P_0 \times V ) \cdot P = 0 $
We can find where the second line intersects this plane, as follows
$ (P_0 \times V) \cdot ( Q_0 + s W ) = 0 $
From this,
$ s = - \dfrac{ Q_0 \cdot ( P_0 \times V ) }{ W \cdot (P_0 \times V ) }$
And this will give us the value of $s$, thus the required Q is
$ Q = Q_0 + s W $
similarly, the value of $t$ where the first line intersects the plane passing through the origin and containing the second line is the solution of the following equation
$ (Q_0 \times W) \cdot ( P_0 + t V ) = 0 $
From this
$ t = - \dfrac{ P_0 \cdot (Q_0 \times W) }{ V \cdot (Q_0 \times W) } $
Now we have $t$ and $s$, so the distance squared between $P$ and $Q$ is
$d^2 = ( P_0 + t V - Q_0 - s W) \cdot (P_0 + t V - Q_0 - s W) $
Numerically, we first compute
$ N_1 = P_0 \times V = (2,1,-1) \times (1,-2,1) = (-1, -3 , -5 ) $
$ N_2 = Q_0 \times W = (8/3, -3, 1 ) \times (2,-1,1) = (-2, -2/3, 10/3 ) $
Therefore,
$ s = - \dfrac{ (8/3, -3, 1) \cdot (-1, -3, -5) }{ (2, -1, 1) \cdot (-1, -3, -5) } = - \dfrac{ -8/3 + 9 - 5 }{-2 + 3 - 5 } = \dfrac{1}{3}  $
And
$ t = - \dfrac{ (2, 1, -1) \cdot (-2, -2/3, 10/3) }{ (1, -2, 1) \cdot (-2, -2/3, 10/3 ) } = - \dfrac{ ( -12 - 2 - 10 ) }{ -6 + 4 + 10 } = \dfrac{24}{8} = 3 $
Now
$ d^2 = \| (2, 1, -1) - (8/3, -3, 1) + 3 (1, -2, 1) - \dfrac{1}{3} (2, -1, 1) \|^2 $
This simplifies to
$ d^2 = \| (5/3 , -5/3 , 2/3)  \|^2 = \dfrac{25+25+4}{9} = \dfrac{ 54 }{9} = 6 $
A: $P=(t+2,-2t+1,t-1)$ and $Q=(2u+\frac{8}{3},-u-3,u+1)$ and $\vec{OP}=k\vec{OQ}$ for some $t,u,k\in\Bbb{R}$. We get the system:
$$t+2=2ku+\frac{8}{3}k$$
$$-2t+1=-ku-3k$$
$$t-1=ku+k$$
Its solution is $k=\frac{3}{2},u=\frac{1}{3},t=3$. Hence, $P=(5,-5,2),Q=(\frac{10}{3}\frac{-10}{3}\frac{4}{3})$ and $PQ^2=(\frac{-5}{3})^2+(\frac{5}{3})^2+(\frac{-2}{3})^2=\frac{54}{9}=6.$
