Let θ be the angle between the planes P _ 1 = → r · ( ^ i + ^ j + 2 ^ k ) = 9 and P _ 2 = → r · (2 ^ i - ^ j + ^ k ) = 15 . Let L be the line that meets P _ 2 at the point (4, - 2, 5) and makes an angle θ with the

Correct answer: 0 Since, θ be the angle between the planes P _ 1 : → r · ( ^ i + ^ j + 2 ^ k ) = 9 and P _ 2 : → r · (2 ^ i - ^ j + ^ k ) = 15 , so cosθ = ( ^ i + ^ j + 2 ^ k ) · (2 ^ i - ^ j + ^ k ) 6 ⇒ cosθ = (3)/(6) = (1)/(2) ⇒ θ = (π)/(3) And, So, α = (π)/(2) - (π)/(3) = (π)/(6) Hence ( ^ 2 θ)( ^ 2 α) = 3 × 3 = 9

JEE Main 202331 JanuaryMorning Shift

Exam: JEE Main
Year: 2023
Date: 31 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Main 2023 · 31 January · Morning Shift · Mathematics · Three Dimensional Geometry

Let θ be the angle between the planes P _ 1 = → r · ( ^ i + ^ j + 2...

Let

\theta

be the angle between the planes

{P}_{1} = \overset{\to}{r} \cdot \left(\overset{^}{i} + \overset{^}{j} + 2\overset{^}{k}\right) = 9 and {P}_{2} = \overset{\to}{r} \cdot \left(2\overset{^}{i} - \overset{^}{j} + \overset{^}{k}\right) = 15

. Let

L

be the line that meets

{P}_{2}

at the point \left(4, - 2, 5\right) and makes an angle

\theta

with the normal of

{P}_{2}

. If

\alpha

is the angle between

L and {P}_{2}

then \left(\tan^{2}\theta\right)

\left(\cot^{2}\alpha\right)

is equal to _____ . Let θ be the angle between the planes P _ 1 = → r · ( ^ i + ^ j + 2 ^ k ) = 9 and P _ 2 = → r · (2 ^ i - ^ j + ^ k ) =

Correct answer

Step-by-step solution

Since, $\theta$ be the angle between the planes ${P}_{1}: \overset{\to}{r} \cdot \left(\overset{^}{i} + \overset{^}{j} + 2\overset{^}{k}\right) = 9 and {P}_{2}: \overset{\to}{r} \cdot \left(2\overset{^}{i} - \overset{^}{j} + \overset{^}{k}\right) = 15$ , so

$cos\theta = \dfrac{\left(\overset{^}{i} + \overset{^}{j} + 2\overset{^}{k}\right) \cdot \left(2\overset{^}{i} - \overset{^}{j} + \overset{^}{k}\right)}{6}$

$⇒ cos\theta = \dfrac{3}{6} = \dfrac{1}{2}$

$⇒ \theta = \dfrac{\pi}{3}$

And,

Let θ be the angle between the planes P _ 1 = → r · ( ^ i + ^ j + 2 ^ k ) = 9 and P _ 2 = → r · (2 ^ i - ^ j + ^ k ) =

So, $\alpha = \dfrac{\pi}{2} - \dfrac{\pi}{3} = \dfrac{\pi}{6}$

Hence

\left(\tan^{2}\theta\right)\left(\cot^{2}\alpha\right) = 3 \times 3 = 9

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