Q51 / 120 · NDA 2024 (Phase 2) · Mathematics

What is $3 \alpha+2 \beta$ equal to if $(2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\alpha \hat{j}+\beta \hat{k})$ is a null vector?

  • Correct answer
    A
    $36$
  • B
    $33$
  • C
    $30$
  • D
    $27$

Solution

Given that, $(2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+27 \hat{\mathrm{k}}) \times(\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+\beta \hat{\mathrm{k}})=0$ $\begin{aligned} & \Rightarrow\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \alpha & \beta \end{array}\right|=0 \\ & \Rightarrow \hat{i}(6 \beta-27 \alpha)-\hat{j}(2 \beta-27)+\hat{k}(2 \alpha-6)=0 \end{aligned}$ Comparing both sides of above equation, we get:- $\begin{aligned} & 6 \beta-27 \alpha=0 \Rightarrow 2 \beta=9 \alpha \\ & 2 \beta-27=0 \Rightarrow \beta=\frac{27}{2} \text { and } 2 \alpha-6=0 \Rightarrow \alpha=3 \end{aligned}$ So, $3 \alpha+2 \beta=3 \times 3+2 \times \frac{27}{2}=9+27=36$

Hence, the correct answer is option: 'A'.

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