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JEE Mathematics 2024 Question with Solution

If λ>0\lambda > 0, and θ\theta is the angle between vectors a⃗=i^+λj^−3k^\vec a = \hat i + \lambda \hat j - 3\hat k and b⃗=3i^−j^+2k^\vec b = 3\hat i - \hat j + 2\hat k, such that a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b are mutually perpendicular, then the value of (14cos⁡(θ))2(14 \cos(\theta))^2 is equal to:

  • A

    2525

  • B

    2020

  • C

    5050

  • D

    4040

Answer

Correct answer:A

Step-by-step solution

the solution appears unrelated

Given: The question is about vectors a⃗=i^+λj^−3k^\vec a = \hat i + \lambda \hat j - 3\hat k and b⃗=3i^−j^+2k^\vec b = 3\hat i - \hat j + 2\hat k, with a⃗+b⃗\vec a + \vec b perpendicular to a⃗−b⃗\vec a - \vec b.

Find: The value of (14cos⁡θ)2(14\cos\theta)^2.

The solution discusses a different problem involving a=sin⁡−1(sin⁡5)a = \sin^{-1}(\sin 5) and b=cos⁡−1(cos⁡5)b = \cos^{-1}(\cos 5), so it is unrelated to this vector question. Because the solution is for a completely different question and the answer key conflicts with the solution, a reliable answer cannot be derived from the provided materials.

Therefore, the answer is marked as AMBIGUOUS.

Common mistakes

  • Using the unrelated the solution about inverse trigonometric functions for this vector question is incorrect because it does not involve λ\lambda, dot products, or the angle between vectors. Instead, use the perpendicularity condition on a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b.

  • A common mistake is to assume perpendicular vectors imply a⃗⋅b⃗=0\vec a \cdot \vec b = 0 directly. Here the condition is (a⃗+b⃗)⋅(a⃗−b⃗)=0(\vec a + \vec b) \cdot (\vec a - \vec b) = 0, which leads to ∣a⃗∣2=∣b⃗∣2|\vec a|^2 = |\vec b|^2, not necessarily a⃗⋅b⃗=0\vec a \cdot \vec b = 0.

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