A line with direction ratios (1,-1,2) intersects the lines L_1: x2=y3=z+13 and L_2: x+1-1=…

Question

A line with direction ratios $(1,-1,2)$ intersects the lines $L_1:\ \dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z+1}{3}$ and $L_2:\ \dfrac{x+1}{-1}=\dfrac{y-2}{1}=\dfrac{z}{4}$ at the points $P$ and $Q$ respectively. If $PQ^2=\dfrac{\alpha}{225}$, then $\alpha$ is equal to:

Accepted Answer

$P=(2\lambda,\ 3\lambda,\ 3\lambda-1)$ on $L_1$; $Q=(-\mu-1,\ \mu+2,\ 4\mu)$ on $L_2$. Since $\vec{PQ}\parallel(1,-1,2)$:
$$\frac{2\lambda+\mu+1}{1}=\frac{3\lambda-\mu-2}{-1}=\frac{3\lambda-4\mu-1}{2}.$$
Solving gives $\lambda=\dfrac15,\ \mu=-\dfrac{8}{15}$.
$$P=\left(\frac25,\frac35,-\frac25\right),\qquad Q=\left(-\frac{7}{15},\frac{22}{15},-\frac{32}{15}\right).$$
$$\vec{PQ}=\left(-\frac{13}{15},\frac{13}{15},-\frac{26}{15}\right).$$
$$PQ^2=\frac{169+169+676}{225}=\frac{1014}{225}.$$
So $\alpha=1014$.

Solutions · 1