Q216 / 270 · NDA 2024 (Phase 1) · Mathematics

Let $\vec{a}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$. If $\vec{a} \times(\vec{b} \times \vec{a})=\alpha \hat{i}-\beta \hat{j}+\gamma \hat{k}$, then what is the value of $\alpha+\beta+\gamma$ ?

  • A
    8
  • B
    7
  • C
    6
  • Correct answer
    D
    1

Solution

We have, $\vec{a}=\hat{i}-\hat{j}+\hat{k}$ and $\begin{aligned} \vec{b}= & \hat{i}+2 \hat{j}-\hat{k} \\ \vec{a} \times(\vec{b} \times \vec{a})= & (\vec{a} \vec{a}) \vec{b}-(\vec{a} \vec{b}) \vec{a} \\ = & {[(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})] \vec{b} } \\ & -[(\hat{i}-\hat{j}+\hat{k}) \cdot(\hat{i}+2 \hat{j}-\hat{k})] \vec{a} \\ = & (1+1+1) \vec{b}-(1-2-1) \vec{a} \\ = & 3(\hat{i}+2 \hat{j}-\hat{k})+2(\hat{i}-\hat{j}+\hat{k})\end{aligned}$ $=5 \hat{i}+4 \hat{j}-\hat{k}$ $\qquad \ldots(i)$ $\vec{a} \times(\vec{b} \times \vec{a})=\alpha \hat{i}-\beta \hat{j}-\gamma \hat{k}$ Comparing with (i) then we get, $\alpha=5, \beta=-4, \gamma=-1$ Hence, $\alpha+\beta+\gamma=5-4-1=0$.

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

More Vector Algebra Questions