Let the lines L_1: r = i+2 j+3 k+(2 i+3 j+4 k) and L_2: r = (4 i+ j)+(5 i+2 j+ k) intersec…

Question

Let the lines $L_1: \vec r = \hat i+2\hat j+3\hat k+\lambda(2\hat i+3\hat j+4\hat k)$ and $L_2: \vec r = (4\hat i+\hat j)+\mu(5\hat i+2\hat j+\hat k)$ intersect at the point $R$. Let $P$ and $Q$ be points on $L_1$ and $L_2$ respectively such that $|PR|=\sqrt{29}$ and $|PQ| = \sqrt{47/3}$. If $P$ lies in the first octant, then $27(QR)^2$ is equal to

Accepted Answer

Equating $L_1,L_2$: $1+2\lambda=4+5\mu$, $2+3\lambda=1+2\mu$, $3+4\lambda=\mu$. Solving: $\mu=-1,\ \lambda=-1$, and this satisfies the third equation, so $R = (1+2\lambda,2+3\lambda,3+4\lambda)|_{\lambda=-1} = (-1,-1,-1)$.

General point on $L_1$: $P=(1+2\lambda,2+3\lambda,3+4\lambda)$. $\vec{PR} = -(1+\lambda)(2,3,4)$, $|PR|^2 = 29(1+\lambda)^2=29 \Rightarrow \lambda=0$ or $-2$. First octant requires $\lambda=0$, so $P=(1,2,3)$.

General point on $L_2$: $Q=(4+5\mu,1+2\mu,\mu)$. $\vec{PQ}=(3+5\mu,-1+2\mu,\mu-3)$, $|PQ|^2 = 30\mu^2+20\mu+19 = \frac{47}{3} \Rightarrow 90\mu^2+60\mu+10=0 \Rightarrow 9\mu^2+6\mu+1=0 \Rightarrow (3\mu+1)^2=0 \Rightarrow \mu=-\frac13$.

$Q = \left(\frac73,\frac13,-\frac13\right)$. $\vec{QR} = R-Q = \left(-\frac{10}{3},-\frac43,-\frac23\right)$, $|QR|^2 = \frac{100+16+4}{9} = \frac{120}{9}$.

$$27(QR)^2 = 27\times\frac{120}{9} = 360$$

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