QJEEAdv16Maf3ee: JEE Advanced 2016 Mathematics - Class 12 JEEAdv16Maf3ee Solution: Consider a pyramid located in the first octant with as origin,…

Let us consider a pyramid OPQRS in the first octant with O as origin as shown in the following figure:


Now, the midpoint of $O Q$ is $T\left(\frac{3}{2}, \frac{3}{2}, 0\right)$ .

Also, it is obvious that the point $S$ is given by $S\left(\frac{3}{2}, \frac{3}{2}, 3\right)$ .

Now, $O T=\sqrt{\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^2+0}=\frac{3}{2} \sqrt{2}$

Therefore, $\cos \theta=\frac{3}{2} \frac{\sqrt{2}}{3}=\frac{1}{\sqrt{2}} \Rightarrow \theta=\frac{\pi}{4}$

where $\theta$ is the angle between $O Q$ and $O S$ , which is calculated as follows:

The ratio of the plane containing triangle $\triangle O Q S$ can be written as

$\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\3 & 3 & 0 \\3 / 2 & 3 / 2 & 3\end{array}\right|=i(9)-i(9)+k(0)=i(i-j)$

Therefore, the ratio is $(1,-1,0)$ .

Now, the equation of the plane is

$\begin{aligned}&1(x-0)-1(y-0)+0(z-0) =0 \\\\&x-y =0\end{aligned}$

Hence, option (B) is correct.

The coordinates of the length of perpendicular from point $P$ to the plane $x-y=0$ is $(3,0,0)$ .

Hence, the length of perpendicular from point $P$ to the plane containing the triangle $O Q S$ is

$\frac{z-0}{\sqrt{2}}=\frac{3}{\sqrt{2}}$

Hence, option (C) is correct.

Now, points $R$ and $S$ , respectively, are $R(0,3,0)$ and $S\left(\frac{3}{2}, \frac{3}{2}, 3\right) \text {. }$

The equation of line $R S$ is

$\begin{aligned}\bar{r} & =3 \hat{j}+\lambda\left(\frac{3}{2} \hat{i}+\left(\frac{3}{2}-3\right) \hat{j}+3 \hat{k}\right) \\\\& =3 \hat{j}+\lambda\left(\frac{3}{2} \hat{i}-\frac{3}{2} \hat{j}+3 \hat{k}\right)\end{aligned}$


Let point A be the feet of perpendicular from O to line RS:

$\begin{aligned}& A\left(3 \hat{j}+\mu\left(\frac{3}{2} \hat{i}-\frac{3}{2} \hat{j}+3 \hat{k}\right)\right) \\\\& \text {Now, } \overrightarrow{O A}=3 \hat{j}+\mu\left(\frac{3}{2} \hat{i}-\frac{3}{2} \hat{j}+3 \hat{k}\right)\end{aligned}$

That is, $(\vec{b}-\vec{a})=\left(\frac{3}{2} \hat{i}-\frac{3}{2} \hat{j}+3 \hat{k}\right)$

Therefore,

$\overrightarrow{O A} \cdot(\vec{b}-\vec{a})=0$

$\begin{aligned}&\left(3 \hat{j}+\mu\left(\frac{3}{2} \hat{i}-\frac{3}{2} \hat{j}+3 \hat{k}\right)\right) \cdot\left(\frac{3}{2} \hat{i}-\frac{3}{2} \hat{j}+3 \hat{k}\right)=0 \\\\&\frac{-9}{2}+\mu\left(\frac{9}{4}+\frac{9}{4}+9\right)=0 \\\\&\frac{-9}{2}+\mu\left(\frac{9}{2}+9\right)=0 \\\\&\Rightarrow-\frac{1}{2}+\mu\left(\frac{3}{2}\right)=0 \\\\&\Rightarrow \mu=\frac{1}{3}\end{aligned}$

Therefore, $A\left(3 \hat{j}+\frac{\hat{i}}{2}-\frac{\hat{j}}{2}+\hat{k}\right)=A\left(\frac{\hat{i}}{2}+\frac{5 \hat{j}}{2}+\hat{k}\right)$ and hence the perpendicular distance from point $O$ to the straight line containing the line $R S$ is

$|\overrightarrow{O A}|=\sqrt{\frac{1}{4}+\frac{25}{4}+\frac{4}{4}}=\sqrt{\frac{30}{4}}=\sqrt{\frac{15}{2}}$

Hence, option (D) is correct.