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Three Dimensional Geometry

Section: Mathematics

Question : 19 of 37

Marks: +1, -0

(x \geq 0, y \geq 0, z \geq 0)

the acute angle between OQ and OS is $\frac{\pi}{3}$
the equation of the plane containing the triangle OQS is $x - y = 0$
the length of the perpendicular from P to the plane containing the triangle OQS is $\frac{3}{\sqrt{2}}$
the perpendicular distance from O to the straight line containing RS is $\sqrt{\frac{15}{2}}$

Solution:

Given OP = OR = 3 and OPQR is a square

\Rightarrow OQ = 3\sqrt{2}

\Rightarrow OT = \frac{3}{\sqrt{2}}

and

ST = 3

Using ΔSOT,

\tan\theta = \frac{ST}{OT} = \sqrt{2} \Rightarrow \theta = \tan^{-1}\sqrt{2}

clearly, equation of plane containing triangle OQS is Y – X = 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

\overset{-}{r} = 3\hat{j} + t\left(\frac{3}{2}\hat{i} - \frac{3}{2}\hat{j} + 3\hat{k}\right)

= \left(\frac{3t}{2}, 3 - \frac{3t}{2}, 3t\right)

Let co-ordinates of M

= \left(\frac{3t}{2}, 3 - \frac{3t}{2}, 3t\right)

\because \overrightarrow{OM} \cdot \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}}

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