The co-ordinates of the perpendicular drawn from the point $2 \hat{i}-\hat{j}+5 \hat{k}$ to the line…

The co-ordinates of the perpendicular drawn from the point $2 \hat{i}-\hat{j}+5 \hat{k}$ to the line $\vec{r}=(11 \hat{i}-2 \hat{j}-8 \hat{k})+\lambda(10 \hat{i}-4 \hat{j}-11 \hat{k})$ are
  1. $(1,-2,3)$
  2. $(1,2,-3)$
  3. $(-1,2,3)$
  4. $(1,2,3)$

Solution

We have $\bar{r}=(11 \hat{i}-2 \hat{j}-8 \hat{k})+\lambda(10 \hat{i}-4 \hat{j}-11 \hat{k})$ So coordinates of any point on this line are $[(10 \lambda+11),(-11 \lambda-2)$, $(-11 \lambda-8)]$ Let $\mathrm{P} \equiv(2,-1,5)$ and let $\underline{\mathrm{M}}$ be foot of perpendicular. $\therefore$ d.r. of PM are $(10 \lambda+9),(-4 \lambda-1),(-11 \lambda-13)$ Sibce $\mathrm{PM}$ is perpendicular to given line, we write $\begin{aligned} & (10 \lambda+9)(10)+(4 \lambda-1)(-4)+(-11 \lambda-13)(-11)=0 \\ & \therefore 100 \lambda+90+16 \lambda+4+121 \lambda+143=0 \Rightarrow 237=-237 \lambda \\ & \Rightarrow \lambda=-1 \\ & M=-10+11,4-2,11-8) \text { i.e. }(1,2,3) \end{aligned}$

Asked in: MHT CET 2021 (24 Sep Shift 1)

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