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QJEEM24Ma9fdd: JEE Main 2024 Physics - Class 11 JEEM24Ma9fdd Solution: The set of all , for which the vectors and are inclined at an…

The set of all α\alpha , for which the vectors a⃗=αti^+6j^−3k^\vec{a}=\alpha t \hat{i}+6 \hat{j}-3 \hat{k} and b⃗=ti^−2j^−2αtk^\vec{b}=t \hat{i}-2 \hat{j}-2 \alpha t \hat{k} are inclined at an obtuse angle for all t∈Rt \in \mathbb{R} , is

Multiple Choice Options

  • A [0,1)[0,1)
  • B (−43,0]\left(-\frac{4}{3}, 0\right]
  • C (−2,0](-2,0]
  • D (−43,1)\left(-\frac{4}{3}, 1\right)

Step-by-Step Verified Solution

Given a⃗=αti^+6j^−3k^\vec{a}=\alpha t \hat{i}+6 \hat{j}-3 \hat{k}

and b⃗=ti^−2j^−2αtk^\vec{b}=t \hat{i}-2 \hat{j}-2 \alpha t \hat{k}

angle between a⃗\vec{a} and b⃗\vec{b} is given by

cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}

We have, \cos \theta < 0(\because angle between a⃗\vec{a} and b⃗\vec{b} is obtuse)

amp;⇒a⃗⋅b⃗lt;0amp;⇒αt2−12+6αtlt;0∀t∈R\begin{aligned}& \Rightarrow \quad \vec{a} \cdot \vec{b}<0 \\& \Rightarrow \alpha t^2-12+6 \alpha t<0 \forall t \in \mathbb{R}\end{aligned}

If α=0\alpha=0 , then -12<0 (condition holds)

If \alpha \neq 0 \Rightarrow \alpha<0\quad \text{.... (i)}

And maximum value of \alpha t^2+6 \alpha t-12<0

amp;⇒−D4alt;0 (where D is discriminant and a=α)amp;⇒36α2+48α4αgt;0amp;⇒αgt;−43amp;∴α∈(−43,0]\begin{aligned}& \Rightarrow \frac{-D}{4 a}<0 \text { (where } D \text { is discriminant and } a=\alpha) \\& \Rightarrow \frac{36 \alpha^2+48 \alpha}{4 \alpha}>0 \\& \Rightarrow \alpha>\frac{-4}{3} \\& \therefore \alpha \in\left(\frac{-4}{3}, 0\right]\end{aligned}

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