line through origin meets two lines at $P$ and $Q,$ then $(PQ)^2$

Question

A line from the origin meet the lines $\displaystyle \frac{x-2}{1} = \frac{y-1}{-2} = \frac{z+1}{1}$ and $\displaystyle \frac{x-\frac{8}{3}}{2} = \frac{y+3}{-1}=\frac{z-1}{1}$

at points $P$ and $Q$ respectively, then $(PQ)^2$ is

Attempt: assuming equation of line is $\displaystyle \frac{x-0}{a} = \frac{y-0}{b} = \frac{z-0}{c}$

in parametric form $\displaystyle \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = \lambda$

so coordinate of $P(a\lambda,b\lambda,c\lambda)$

in parametric form $\displaystyle \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = \mu$

so coordinate of $Q(a\mu,b\mu,c\mu)$

so $(PQ)^2 = (a^2+b^2+c^2)(\lambda-\mu)^2$

wan,t be able to go further, could some help me, thanks

Answer 1

We then have substituting the point $P $ into the equation of the first line as: $$\frac {a\lambda -2}{1} = \frac {b\lambda + 1}{-2} = \frac {c\lambda +1}{1} = k $$ giving us $$ a\lambda = 2+k$$ $$b\lambda =-2k-1$$ $$c\lambda = k-1$$

Similarly substituting $Q $ in the second line gives us: $$a\mu = 2k_1 + \frac {8}{3} $$ $$b\mu = -k_1-3$$ $$c\mu = k_1 +1$$

We thus get from these equations $$\frac {2+k}{\lambda} = \frac {2k_1 + \frac {8}{3}}{\mu} $$ $$\frac {-2k-1}{\lambda} = \frac {-k_1-3}{\mu} $$ $$\frac {k-1}{\lambda} = \frac {k_1+1}{\mu} $$

Now try to express $\lambda $ in terms of $\mu $ and thus both the points $P $ and $Q $ will be expressed in the same parameter $\mu $. Then find the square of the distance. Hope it helps.