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JEE Mathematics 2026 Question with Solution

Let a vector a⃗=2 i^−j^+λk^, λ>0,\vec a=\sqrt{2}\,\hat i-\hat j+\lambda \hat k,\ \lambda>0, make an obtuse angle with the vector b⃗=−λ2i^+42j^+42k^\vec b=-\lambda^2\hat i+4\sqrt{2}\hat j+4\sqrt{2}\hat k and an angle θ\theta, π6<θ<π2\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 _____.

Answer

Correct answer:5

Step-by-step solution

Standard Method

Given: a⃗=2 i^−j^+λk^\vec a=\sqrt{2}\,\hat i-\hat j+\lambda \hat k with λ>0\lambda>0, and b⃗=−λ2i^+42j^+42k^\vec b=-\lambda^2\hat i+4\sqrt{2}\hat j+4\sqrt{2}\hat k.

Find: The value of α+β+γ\alpha+\beta+\gamma if the set of possible values of λ\lambda is (α,β)−{γ}(\alpha,\beta)-\{\gamma\}.

For the angle between a⃗\vec a and b⃗\vec b to be obtuse, their dot product must be negative:

a⃗⋅b⃗<0\vec a\cdot\vec b<0 (2)(−λ2)+(−1)(42)+λ(42)<0(\sqrt2)(-\lambda^2)+(-1)(4\sqrt2)+\lambda(4\sqrt2)<0 −2λ2−42+42λ<0-\sqrt2\lambda^2-4\sqrt2+4\sqrt2\lambda<0

Dividing by 2\sqrt2,

−λ2+4λ−4<0-\lambda^2+4\lambda-4<0 (λ−2)2>0(\lambda-2)^2>0

Hence,

λ≠2\lambda\ne 2

Now use the angle θ\theta made by a⃗\vec a with the positive zz-axis. Since the direction cosine with the positive zz-axis is az∣a⃗∣\frac{a_z}{|\vec a|},

cos⁡θ=λ2+1+λ2=λλ2+3\cos\theta=\frac{\lambda}{\sqrt{2+1+\lambda^2}}=\frac{\lambda}{\sqrt{\lambda^2+3}}

Given π6<θ<π2\frac{\pi}{6}<\theta<\frac{\pi}{2}, we get

0<cos⁡θ<320<\cos\theta<\frac{\sqrt3}{2}

So,

0<λλ2+3<320<\frac{\lambda}{\sqrt{\lambda^2+3}}<\frac{\sqrt3}{2}

Squaring the inequality,

λ2λ2+3<34\frac{\lambda^2}{\lambda^2+3}<\frac{3}{4} 4λ2<3λ2+94\lambda^2<3\lambda^2+9 λ2<9\lambda^2<9

Since λ>0\lambda>0,

0<λ<30<\lambda<3

Combining both conditions,

λ∈(0,3)−{2}\lambda\in(0,3)-\{2\}

Therefore,

α=0,β=3,γ=2\alpha=0,\quad \beta=3,\quad \gamma=2 α+β+γ=5\alpha+\beta+\gamma=5

Therefore, the required numerical value is 55.

Why the interval becomes an open interval with one excluded point

Given: The two conditions are an obtuse-angle condition and an angular bound with the positive zz-axis.

Find: How these conditions shape the set of allowed values of λ\lambda.

The obtuse-angle condition gives

a⃗⋅b⃗<0\vec a\cdot\vec b<0

which reduces to

(λ−2)2>0(\lambda-2)^2>0

This is true for every real value except

λ=2\lambda=2

So this condition removes exactly one point.

The condition π6<θ<π2\frac{\pi}{6}<\theta<\frac{\pi}{2} implies that θ\theta lies strictly between these two angles, so the corresponding cosine values are also strict:

0<cos⁡θ<320<\cos\theta<\frac{\sqrt3}{2}

That produces the strict interval

0<λ<30<\lambda<3

Thus one condition gives an open interval, and the other removes one interior point from it. Hence the final set is

(0,3)−{2}(0,3)-\{2\}

So,

α=0, β=3, γ=2\alpha=0,\ \beta=3,\ \gamma=2

and therefore the required sum is 55.

Common mistakes

  • Using the obtuse-angle condition as a⃗⋅b⃗>0\vec a\cdot\vec b>0. This reverses the meaning of an obtuse angle. For an obtuse angle, the cosine is negative, so the correct condition is a⃗⋅b⃗<0\vec a\cdot\vec b<0.

  • Forgetting that θ\theta is the angle with the positive zz-axis, so cos⁡θ\cos\theta must be written as the zz-component of a⃗\vec a divided by ∣a⃗∣|\vec a|. Do not use the whole vector or a wrong component; use cos⁡θ=λλ2+3\cos\theta=\frac{\lambda}{\sqrt{\lambda^2+3}}.

  • After obtaining (λ−2)2>0(\lambda-2)^2>0, concluding that all values are allowed without excluding λ=2\lambda=2. A square is zero exactly at one point, so the inequality removes that single value.

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