Question:
The foot of the perpendicular drawn from the point $(4,2,3)$ to the line joining the points
$(1,-2,3)$ and $(1,1,0)$ lies on the plane :
Correct Option: , 4
Solution:
Equation of $\mathrm{AB}=\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}})+\lambda(3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})$
Let coordinates of M $=(1,(1+3 \lambda),-3 \lambda)$.
$\overrightarrow{\mathrm{PM}}=-3 \hat{\mathrm{i}}+(3 \lambda-1) \hat{\mathrm{j}}-3(\lambda+1) \hat{\mathrm{k}}$
$\overline{\mathrm{AB}}=3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$
$\Rightarrow \lambda=-\frac{1}{3}$
$\therefore \quad M=(1,0,1)$
Clearly $\mathrm{M}$ lies on $2 \mathrm{x}+\mathrm{y}-\mathrm{z}=1$.