Exam

JEE Advanced

Year

2012

Paper

Paper 1

Subject

Mathematics

Chapter

Three Dimensional Geometry

JEE Advanced 2012 · Paper 1 · Mathematics · Three Dimensional Geometry

The point P is the intersection of the straight line joining the...

The point $P$ is the intersection of the straight line joining the points $Q(2,3,5)$ and $R(1,-1,4)$ with the plane $5 x-4 y-$ $z=1$. If $S$ is the foot of the perpendicular drawn from the point $T(2,1,4)$ to $Q R$, then the length of the line segment $P S$ is

Step-by-step solution

Equation of st. line joining $Q(2,3,5)$ and $R(1,-1,4)$ is $\frac{x-2}{-1}=\frac{y-3}{-4}=\frac{z-5}{1}=\lambda$ Let $P(-\lambda+2,-4 \lambda+3,-\lambda+5)$ Since $P$ also lies on $5 x-4 y-z=1$ $\therefore-5 \lambda+10+16 \lambda-12+\lambda-5=1$ $\Rightarrow 12 \lambda=8 \Rightarrow \lambda=\frac{2}{3} \quad \therefore P=$\left(\frac{4}{3}, \frac{1}{3}, \frac{13}{3}\right)$Now let another point$S$on$Q R$be$(-\mu+2,-4 \mu+3,-\mu+5)$Since$S$is the foot of perpendicular drawn from$T(2,1,4)$to$Q R$, where dr's of$S T$are$\mu, 4 \mu-2, \mu-1$and dr's of$Q R$are$-1,-4,-1 $$\begin{array}{ll} \therefore & -\mu-16 \mu+8-\mu+1=0 \Rightarrow 18 \mu=9 \Rightarrow \mu=\frac{1}{2} \\ \therefore & S=\left(\frac{3}{2}, 1, \frac{9}{2}\right) \end{array}$$ \therefore \quad$Distance between$P$and$S $$\begin{array}{l} =\sqrt{\left(\frac{4}{3}-\frac{3}{2}\right)^{2}+\left(\frac{1}{3}-1\right)^{2}+\left(\frac{13}{3}-\frac{9}{2}\right)^{2}} \\ =\sqrt{\frac{1}{36}+\frac{4}{9}+\frac{1}{36}}=\frac{1}{\sqrt{2}} \end{array}$$