What is 3α + 2β equal to if (2î + 6ĵ + 27k̂) × (î + αĵ + βk̂) is a null vector?

Question

Question

This question was previously asked in

NDA-II 2024 (Maths) Official Paper (Held On: 01 Sept, 2024)

Answer 1

Explanation:

Given:

⇒ (2î + 6ĵ + 27k̂) × (î + αĵ + βk̂) = 0

⇒ \(\begin{bmatrix} \hat{i} &\hat{j}&\hat{k}\\\ 2&6&27\\\ 1&α&β\end{bmatrix}\ \) = 0

⇒ \(\hat{i} (6β -27α) - \hat{j}(2β-27)+\hat{k}(2α-6) =0\)

Comparing both sides, we get

6β – 27α = 0

⇒ 2β = 9α

2β - 27 = 0

⇒ β = 27/2

Also

⇒ 2α – 6 = 0

α =3

Now,

3α + 2β = \(3\times 3 + 2 \times \frac{27}{2}\) =36

∴ Option (1) is correct

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