Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

$OQ=3i+3jOS=\frac{3}{2}i+\frac{3}{2}j+3k\therefore \cos \theta =\frac{1}{\sqrt{3}}$ As, $\mathrm{O},\mathrm{Q},\mathrm{S}$ lies on $x-y=0$, the plane containing triangle OQS is $x-y=0$. Length of the perpendicular from $\mathrm{P}(3,0,0)$ to the plane containing $OQS=\frac{|3-0|}{\sqrt{1+1}}=\frac{3}{\sqrt{2}}$ Suppose foot of the perpendicular from O to RS be M , and let M divide RS in the ration $\mathrm{k}:1$. Coordinate of M are $\left(\frac{3k/2}{k+1},\frac{(3k/2)+3}{k+1},\frac{3k}{k+1}\right)$ Direction ratio of OM are $k,k+2,2k$ Direction ratio of RS : $\frac{3}{2},\frac{-3}{2},3$ or $1,-1,2$ As $OM\perp RS$ we get $k-(k+2)+4k=0\Rightarrow k=\frac{1}{2}$ $\therefore M=\left(\frac{1}{2},\frac{5}{2},1\right)$ and $OM=\sqrt{\frac{1}{4}+\frac{25}{4}+1}=\frac{\sqrt{15}}{2}$