Question
A line with direction ratios $1, - 1,2$ intersects the lines $\frac{x}{2} = \frac{y}{3} = \frac{z + 1}{3}$ and $\frac{x + 1}{- 1} = \frac{y - 2}{1} = \frac{z}{4}$ at the points P and Q , respectively. If the length of the line segment PQ is $\alpha$, then $225\alpha^{2}$ is equal to:
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Solution
Sol. [Key Concept: Write general points P, Q on the two lines; impose direction ratios $1, - 1,2$ and find $|PQ|$.]
$P(2\lambda,3\lambda,3\lambda - 1)$ and $Q( - \mu - 1,\mu + 2,4\mu)$
D.R of PQ $2\lambda + \mu + 1,3\lambda - \mu - 2,3\lambda - 4\mu - 1$, $\frac{2\lambda + \mu + 1}{1} = \frac{3\lambda - \mu - 2}{- 1} = \frac{3\lambda - 4\mu - 1}{2}$
Solving $\lambda = \frac{1}{5}$ and $\mu = \frac{- 8}{15}$
$${P\left( \frac{2}{5},\frac{3}{5},\frac{- 2}{5} \right)\& Q\left( \frac{- 7}{15},\frac{22}{15},\frac{- 32}{15} \right) }{{PQ}^{2} = \left( \frac{13}{15} \right)^{2} + \left( \frac{13}{15} \right)^{2} + \left( \frac{26}{15} \right)^{2} }{\alpha^{2} = \frac{1014}{225}\therefore 225\alpha^{2} = 1014 }$$
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