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JEE Main 202622 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Vector Algebra

JEE Main 2026 · 22 January · Evening Shift · Mathematics · Vector Algebra

Let a vector a = 2 i - j +λ k , λ>0 , make an obtuse angle with the...

Let a vector a→=2i^−j^+λk^,λ>0\overrightarrow{\mathrm{a}}=\sqrt{2} \hat{i}-\hat{j}+\lambda \hat{k}, \lambda>0, make an obtuse angle with the vector b→=−λ2i^+42j^+42k^\overrightarrow{\mathrm{b}}=-\lambda^{2} \hat{i}+4 \sqrt{2} \hat{j}+4 \sqrt{2} \hat{k} and an angle θ,π6<θ<π2\theta, \frac{\pi}{6}<\theta<\frac{\pi}{2}, with the positive zz-axis. If the set of all possible values of λ\lambda is (α,β)−γ(\alpha, \beta)-\gamma, then α+β+γ\alpha+\beta+\gamma is equal to ____\_\_\_\_。

Correct answer

0

Step-by-step solution

a→.k^∣a→∣=cos⁡θ⇒λ3+λ2=cos⁡θ⇒0<λ3+λ2<32⇒λ>0&4λ2<9+3λ2⇒λ2<9⇒λ∈(0,3)….(1)⇒a→⋅b→<0⇒−2λ2−42+42λ<0⇒λ2−4λ+4>0⇒(λ−2)2>0⇒λ≠2 from (1)&(2)λ∈(0,3)−{2}∴α=0,β=3,γ=2⇒α+β+γ=5\frac{\overrightarrow{\mathrm{a}} . \hat{\mathrm{k}}}{|\overrightarrow{\mathrm{a}}|}=\cos \theta \Rightarrow \frac{\lambda}{\sqrt{3+\lambda^2}}=\cos \theta \Rightarrow 0<\frac{\lambda}{\sqrt{3+\lambda^2}}<\frac{\sqrt{3}}{2} \Rightarrow \lambda>0 \& 4 \lambda^2<9+3 \lambda^2 \Rightarrow \lambda^2<9 \Rightarrow \lambda \in(0,3) \quad \quad \ldots .(1) \Rightarrow \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{ b}}<0 \Rightarrow-\sqrt{2} \lambda^2-4 \sqrt{2}+4 \sqrt{2} \lambda<0 \Rightarrow \lambda^2-4 \lambda+4>0 \Rightarrow(\lambda-2)^2>0 \Rightarrow \lambda \neq 2 \text{ from }(1) \&(2) \lambda \in(0,3)-\{2\} \therefore \alpha=0, \beta=3, \gamma=2 \Rightarrow \alpha+\beta+\gamma=5
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