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Q. 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:

JEE MainJEE Main 2020Three Dimensional Geometry

Solution:

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Equation of $AB =\overrightarrow{ r }=(\hat{ i }+\hat{ j })+\lambda(3 \hat{ j }-3 \hat{ k })$
Let coordinates of $M =(1,(1+3 \lambda),-3 \lambda)$.
$\overrightarrow{ PM }=-3 \hat{ i }+(3 \lambda-1) \hat{ j }-3(\lambda+1) \hat{ k }$
$\overrightarrow{ AB }=3 \hat{ j }-3 \hat{ k }$
$\because \overrightarrow{PM} \perp \overrightarrow{AB}$
$\Rightarrow \overrightarrow{PM} . \overrightarrow{AB} =0$
$\Rightarrow 3(3 \lambda-1)+9(\lambda+1)=0$
$\Rightarrow \lambda=-\frac{1}{3}$
$\therefore M =(1,0,1)$
Clearly M lies on $2 x+y-z=1$