JEE Advanced 2016 · Paper 1 · Mathematics Q42

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MathematicsMultiple correct+4 / -2Question 42 of 54

Consider a pyramid $OPQRS$ located in the first octant $(x\ge0,\ y\ge0,\ z\ge0)$ with $O$ as origin, and $OP$ and $OR$ along the $x$ -axis and the $y$ -axis, respectively. The base $OPQR$ of the pyramid is a square with $OP=3$ . The point $S$ is directly above the mid-point $T$ of diagonal $OQ$ such that $TS=3$ . Then

the acute angle between

OQ

and

OS

\dfrac{\pi}{3}

the equation of the plane containing the triangle

OQS

x-y=0

the length of the perpendicular from

P

to the plane containing the triangle

OQS

\dfrac{3}{\sqrt2}

the perpendicular distance from

O

to the straight line containing

RS

\sqrt{\dfrac{15}{2}}

Answer and worked solution

Official answer: B, C, D

Coordinates. $O(0,0,0)$ , $P(3,0,0)$ , $Q(3,3,0)$ , $R(0,3,0)$ , $T\left(\frac32,\frac32,0\right)$ and $S\left(\frac32,\frac32,3\right)$ .

(A) $\cos\angle(OQ,OS)=\dfrac{9}{3\sqrt2\cdot\sqrt{27/2}}=\dfrac1{\sqrt3}$ , so the angle is not $\frac\pi3$ . False.
(B) $O$ , $Q$ and $S$ all satisfy $x=y$ . The plane is $x-y=0$ . True.
(C) Distance from $P$ to $x-y=0$ : $\dfrac{3}{\sqrt2}$ . True.
(D) Line RS has direction $(1,-1,2)$ . Then
$d=\frac{|\overrightarrow{OR}\times(1,-1,2)|}{\sqrt6}=\frac{|(6,0,-3)|}{\sqrt6}=\sqrt{\frac{45}6}=\sqrt{\frac{15}2}.$
True.

(B), (C), (D)

Topics:Angles and distances Equation of a plane

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JEE Advanced 2016 · Paper 1, Mathematics question 42