The length of the perpendicular from the point $\mathrm{A}(1,-2,-3)$ on the line…

The length of the perpendicular from the point $\mathrm{A}(1,-2,-3)$ on the line $\frac{x-1}{2}=\frac{y+3}{-1}=\frac{z+1}{-2}$ is
  1. 6 units
  2. 3 units
  3. 2 units
  4. 4 units

Solution

Let $\frac{x-1}{2}=\frac{y+3}{-1}=\frac{z+1}{-2}=\lambda$ $\therefore \quad$ Any general point on this line is $\mathrm{Q}(2 \lambda+1,-\lambda-3,-2 \lambda-1)$
The direction ratios of AQ are $2 \lambda,-\lambda-1,-2 \lambda+2$
Since $A Q$ is perpendicular to given lines. $\begin{aligned} & 2(2 \lambda)-1(-\lambda-1)-2(-2 \lambda+2)=0 \\ \Rightarrow & 4 \lambda+\lambda+1+4 \lambda-4=0 \\ & \Rightarrow 9 \lambda-3=0 \\ \Rightarrow & \lambda=\frac{1}{3} \\ \therefore \quad & \mathrm{Q} \equiv\left(\frac{5}{3}, \frac{-10}{3}, \frac{-5}{3}\right) \end{aligned}$ $\therefore \quad \mathrm{AQ}=\sqrt{\left(1-\frac{5}{3}\right)^2-\left(-2+\frac{10}{3}\right)^2+\left(-3+\frac{5}{3}\right)^2}$ $\begin{aligned} & =\sqrt{\frac{4}{9}+\frac{16}{9}+\frac{16}{9}}=\sqrt{\frac{36}{9}}=\sqrt{4} \\ \therefore \quad A Q & =2 \text { units }\end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 1)

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