Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

$\mathrm{O}\left(\mathrm{0,0},0\right)\to \mathrm{Origin}$
$\mathrm{P}\left(\mathrm{3,0},0\right)\to \mathrm{on\; x\; axis}$
$$ \mathrm{R}\left(\mathrm{0,3},0\right)\to \mathrm{on\; y\; axis}$$
$\mathrm{Q}\left(\mathrm{3,3},0\right)$
$\mathit{T}\left(\frac{3}{2},\frac{3}{2},0\right),5\left(\frac{3}{2},\frac{3}{2},3\right)$


Given OP = OR = 3 and OPQR is a square
$\Rightarrow$$OQ=3\sqrt{2} \Rightarrow OT=\frac{3}{\sqrt{2}}$and$ST=3$
Let θ be a angle between OQ & OS
Using$\mathrm{\Delta }SOT, \tan \theta =\frac{ST}{OT}= \sqrt{2} \Rightarrow \theta ={\tan }^{-1}\sqrt{2}$

Clearly, equation of plane containing triangle OQS is x - y = 0   as$\mathrm{O}\left(\mathrm{0,0},0\right),\mathrm{Q}\left(\mathrm{3,3},0\right),\mathrm{S}\left(\frac{3}{2},\frac{3}{2}\mathrm{,\; 3}\right)\mathrm{ \; lies\; on\; it}$
$\mathrm{let\; ax}+\mathrm{by}+\mathrm{cz}=\mathrm{d}$
$\left(\mathrm{0,0},0\right) \Rightarrow d=0$
$\left(\mathrm{3,3},0\right) \Rightarrow a+b=0$
$\left(\frac{3}{2},\frac{3}{2},3\right)\Rightarrow \frac{3a}{2}+\frac{3b}{2}+3c=0$
$\Rightarrow c=0$
$\Rightarrow b=-a$
$x-y=0$
Also, length of perpendicular from P to the plane containing the triangle OQS is PT$=\frac{3}{\sqrt{2}}$.
Also equation of RS is$\overrightarrow{r}=3\widehat{j}+t \left(\frac{3}{2}\widehat{i}-\frac{3}{2}\widehat{j}+3\widehat{k}\right)$
$=\left(\frac{3t}{2}, 3-\frac{3t}{2}, 3t\right)$

Let M be foot of$\perp$from origin on line passing through R,S
Let co - ordinates of$M=\left(\frac{3t}{2}, 3-\frac{3t}{2}, 3t\right)$
$\because$$\overrightarrow{OM} . \overrightarrow{RS}=0$
$\Rightarrow$$\frac{9}{4}t-\frac{3}{2}\left(3-\frac{3t}{2}\right)+9t=0$
$\Rightarrow$$\frac{9t}{2}+9t=\frac{9}{2}$
$\Rightarrow$$t=\frac{1}{3}$
$\therefore$$M=\left(\frac{1}{2},\frac{5}{2}, 1\right)$
$\Rightarrow$$OM= \sqrt{\frac{1}{4}+\frac{25}{4}+1}= \sqrt{\frac{30}{4}}= \sqrt{\frac{15}{2}}$