A line having direction ratios $1,-4,2$ intersects the lines $\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}$…

A line having direction ratios $1,-4,2$ intersects the lines $\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}$ and $\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}$ at the points $A$ and $B$ resp., then co-ordinates of points A and B are
  1. $\begin{aligned} & \mathrm{A}(-8,6,-7) \\ & \mathrm{B}(-6,-2,-3)\end{aligned}$
  2. $\begin{aligned} & \mathrm{A}(8,6,7) \\ & \mathrm{B}(6,2,3)\end{aligned}$
  3. $\begin{aligned} & \mathrm{A}(8,6,7) \\ & \mathrm{B}(6,-2,-3)\end{aligned}$
  4. $\begin{aligned} & \mathrm{A}(7 ; 6,8) \\ & \mathrm{B}(-3,-2,6)\end{aligned}$

Solution

Let $\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}=\lambda$ $\Rightarrow x=3 \lambda+7, y=-\lambda+1, z=\lambda-2$
Let $\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}=\mu$ $\Rightarrow x=2 \mu, y=3 \mu+7, z=\mu$
Co-ordinates of a point on the first line are $\mathrm{A}(3 \lambda+7,1-\lambda, \lambda-2)$
Co-ordinates of a point on the second line are $\mathrm{B}(2 \mu, 3 \mu+7, \mu)$ D.r.s. of $A B$ are $3 \lambda-2 \mu+7,-\lambda-3 \mu-6, \lambda-\mu-2$ D.r.s. of $A B$ are $1,-4,2$ $\begin{aligned} \therefore \quad & \frac{3 \lambda-2 \mu+7}{1}=\frac{-\lambda-3 \mu-6}{-4}=\frac{\lambda-\mu-2}{2} \\ & 3 \lambda-2 \mu+7=\frac{\lambda+3 \mu+6}{4} \\ & \Rightarrow \lambda-\mu+2=0 ...(i)\\ & \frac{\lambda+3 \mu+6}{4}=\frac{\lambda-\mu-2}{2} \\ & \Rightarrow \lambda-5 \mu-10=0...(ii) \end{aligned}$
Solving (i) and (ii), we get $\begin{aligned} \lambda & =-5, \mu=-3 \\ \therefore \quad & A \equiv(-8,6,-7), \\ & B \equiv(-6,-2,-3) \end{aligned}$

Asked in: MHT CET 2024 (16 May Shift 2)

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