The set of all α , for which the vectors a =α t i +6 j -3 k and b =t i -2 j -2 α t k are inclined at an obtuse angle for all t ∈ ℝ , is

Correct answer: C. (-(4)/(3), 0] aligned & a =α i +6 j -3 k & b =t i -2 j -2 α k & so a b < 0, t ∈ R & α t^2-12+6 α t < 0 & α t^2+6 α t-12 < 0, t ∈ R & α < 0 , and D < 0 & 36 α^2+48 α < 0 & 12 α(3 α+4) < 0 & (-4)/(3) < α < 0 aligned also for a=0, a · b < 0 hence a α ∈((-4)/(3) , 0]

JEE Main 20248 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 8 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2024 · 8 April · Morning Shift · Mathematics · Vector Algebra

The set of all α , for which the vectors a =α t i +6 j -3 k and b...

The set of all

\alpha

, for which the vectors

\vec{a}=\alpha t \hat{i}+6 \hat{j}-3 \hat{k}

and

\vec{b}=t \hat{i}-2 \hat{j}-2 \alpha t \hat{k}

are inclined at an obtuse angle for all

t \in \mathbb{R}

, is

Options

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

Correct answer

C. $\left(-\frac{4}{3}, 0\right]$

Step-by-step solution

\begin{aligned} & \overrightarrow{\mathrm{a}}=\alpha \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{b}}=\mathrm{t} \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-2 \alpha \hat{\mathrm{k}} \\ & \text{ so } \overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} < 0, \forall \mathrm{t} \in \mathrm{R} \\ & \alpha \mathrm{t}^2-12+6 \alpha \mathrm{t} < 0 \\ & \alpha \mathrm{t}^2+6 \alpha \mathrm{t}-12 < 0, \forall \mathrm{t} \in \mathrm{R} \\ & \alpha < 0 \text{,\text{ and } } \mathrm{D} < 0 \\ & 36 \alpha^2+48 \alpha < 0 \\ & 12 \alpha(3 \alpha+4) < 0 \\ & \frac{-4}{3} < \alpha < 0\end{aligned}

also for

\mathrm{a}=0, \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}} < 0

hence a

\alpha \in\left(\frac{-4}{3}

, 0\right]

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