JEE Main 202330 JanuaryMorning Shift

Exam: JEE Main
Year: 2023
Date: 30 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Main 2023 · 30 January · Morning Shift · Mathematics · Three Dimensional Geometry

Let a unit vector → OP make angle α, β, γ with the positive...

Let a unit vector

\overset{\to}{OP}

make angle

\alpha, \beta, \gamma

with the positive directions of the co-ordinate axes

OX, OY

OZ

respectively, where

\beta \in \left(0, \dfrac{\pi}{2}\right)

\overset{\to}{OP}

is perpendicular to the plane through points \left(1, 2, 3\right), \left(2, 3, 4\right)

and \left(1, 5, 7\right)

, then which one of the following is true ?

Options

A $\alpha \in \left(\dfrac{\pi}{2}, \pi\right)$ and $\gamma \in \left(\dfrac{\pi}{2}, \pi\right)$
B $\alpha \in \left(0, \dfrac{\pi}{2}\right)$ and $\gamma \in \left(0, \dfrac{\pi}{2}\right)$
C $\alpha \in \left(\dfrac{\pi}{2}, \pi\right)$ and $\gamma \in \left(0, \dfrac{\pi}{2}\right)$
D $\alpha \in \left(0, \dfrac{\pi}{2}\right)$ and $\gamma \in \left(\dfrac{\pi}{2}, \pi\right)$

Correct answer

A. $\alpha \in \left(\dfrac{\pi}{2}, \pi\right)$ and $\gamma \in \left(\dfrac{\pi}{2}, \pi\right)$

Step-by-step solution

Equation of plane passing through the points $\left(1, 2, 3\right)$ , \left(2, 3, 4\right) $and \left(1, 5, 7\right)$ is

\left| $\begin{cases}x - 1 & y - 2 & z - 3 \\ 1 & 1 & 1 \\ 0 & 3 & 4\end{cases}$ \right| = 0 $$ ⇒ \left(x - 1\right) - 4\left(y - 2\right) + 3\left(z - 3\right) = 0 $$ ⇒ x - 4y + 3z = 2 $Direction ratios of normal of the plane is Direction cosines of normal of the plane is $ \left<\pm \dfrac{1}{\sqrt{26}}, ∓ \dfrac{4}{\sqrt{26}}, \pm \dfrac{3}{\sqrt{26}}\right> $Hence,$ cos\beta = \dfrac{4}{\sqrt{26}} > 0 ⇒ \beta \in \left(0, \dfrac{\pi}{2}\right) $,$ cos\alpha = \dfrac{- 1}{\sqrt{26}} ⇒ \alpha \in \left(\dfrac{\pi}{2}, \pi\right) $,$ cos\gamma = - \dfrac{3}{\sqrt{26}} ⇒ \gamma \in \left(\dfrac{\pi}{2}, \pi\right)

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