Let a vector \(\vec{a}=\sqrt{2} \hat{i}-\hat{j}+\lambda \hat{k}, \lambda>0\), make an obtuse angle with the vector \(\vec{b}=-\lambda^{2} \hat{i}+4 \sqrt{2} \hat{j}+4 \sqrt{2} \hat{k}\) and an angle \(\theta, \frac{\pi}{6}<\theta< \frac{\pi}{2}\), with the positive \(z\)-axis. If the set of all possible values of \(\lambda\) is \((\alpha, \beta)-\{\gamma\}\), then \(\alpha+\beta+\gamma\) is equal to ______.

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Vector Algebra

Let a vector

a = 2 \hat{i} - \hat{j} + λ \hat{k}, λ > 0

, make an obtuse angle with the vector

b = - λ^{2} \hat{i} + 42 \hat{j} + 42 \hat{k}

and an angle

θ, \frac{π}{6} < θ < \frac{π}{2}

, with the positive

z

-axis. If the set of all possible values of

λ

(α, β) - {γ}

, then

α + β + γ

is equal to ______.

Answer: 5

The angle with the positive

z

-axis gives

cos θ = \frac{λ}{3 + λ ^{2}}

. Since

\frac{π}{6} < θ < \frac{π}{2}

means

0 < cos θ < \frac{3}{2}

, we get

\frac{λ ^{2}}{3 + λ ^{2}} < \frac{3}{4}

, i.e.

λ^{2} < 9

, so

λ \in (0, 3)

. For an obtuse angle with

b

a \cdot b = - 2 λ^{2} - 42 + 42 λ = - 2 (λ - 2)^{2} < 0,

which holds for every

λ \neq = 2

. Thus the possible values form

(0, 3) - {2}

, so

α = 0, β = 3, γ = 2

and

α + β + γ = 5

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Integer answer