question_answer

The point P is the intersection of the straight line joining the points 0(2, 3, 5) and R(1, -1, 4) with the plane . If S is the foot of the perpendicular drawn from the point  to QR, then the length of the line segment PS is

A)                              

B)  

C) 2                                 

D)

Correct Answer: A

Solution :

Direction ratios of QR are 1, 4, 1 Equation of line QR is \[\frac{x-2}{1}=\frac{y-3}{4}=\frac{z-5}{1}=\lambda \] Let \[\operatorname{P}\equiv \left( 2+\lambda ,\,3+4\,\lambda ,\,\,5+\lambda  \right)\] \[10+5\lambda -12-16\lambda -5-\lambda =1\,\,\,\,\, \left[ From\,\,5x-4y-z=1 \right]\]\[-7-12\lambda =1\,\,\,\Rightarrow \,\,\,\lambda =\frac{-2}{3}\] Then \[P\equiv \left( \frac{4}{3},\frac{1}{3},\frac{13}{3} \right)\] Let \[S=\,\,(2+\mu ,\,\,3+4\mu ,\,\,5+\mu )\] \[\vec{T}S=\,\,(\mu )\hat{i}+(4\mu +2)\hat{j}\,+\,(\mu +1)\hat{k}\] \[\vec{T}S=\,\,(\hat{i}+4\hat{j}+\hat{k})=0\,\,\Rightarrow \,\,\mu +16\mu +8+\mu +1=0\] \[\mu =-\frac{1}{2}\Rightarrow S=\left( \frac{3}{2},\,\,1,\,\,\frac{9}{2} \right)\] \[PS=\sqrt{{{\left( \frac{4}{3}-\frac{3}{2} \right)}^{3}}+{{\left( \frac{1}{3}-1 \right)}^{2}}+{{\left( \frac{13}{3}-\frac{9}{2} \right)}^{2}}}\,\,=\,\sqrt{\frac{1}{36}+\frac{4}{9}+\frac{1}{36}}\] \[=\,\,\sqrt{\frac{1}{18}+\frac{4}{9}}=\sqrt{\frac{9}{18}}\,\,=\,\,\frac{1}{\sqrt{2}}\]