Q215 / 270 · NDA 2023 (Phase 1) · Mathematics
Paragraph:
Let $\vec{a}=3 \hat{i}+3 \hat{j}+3 \hat{k}$ and $\vec{c}=\hat{j}-k$. Let $\vec{b}$ be such that $\vec{a} \cdot \vec{b}=27$ and $\vec{a} \times \vec{b}=\overrightarrow{9 c}$
Question:
What is $\vec{b}$ equal to?
Question: What is $\vec{b}$ equal to?
- A$3 \hat{i}+4 \hat{j}+2 \hat{k}$
- Correct answerB$5 \hat{i}+2 \hat{j}+2 \hat{k}$
- C$5 \hat{i}-2 \hat{j}+6 \hat{k}$
- D$3 \hat{i}+3 \hat{j}+4 \hat{k}$
Solution
$\vec{a}=3 \hat{i}+3 \hat{j}+3 \hat{k}$
$\vec{c}=\hat{j}-\hat{k}$
$\vec{a} \cdot \vec{b}=27$
$\vec{a} \times \vec{b}=9 \vec{c}$
Let $b=x \hat{i}+y \hat{j}+z \hat{k}$
$\vec{a} \cdot \vec{b}=3 x+3 y+3 z=27$
$\Rightarrow x+y+z=9$
Also, $\vec{a} \times b=9 \vec{c}$
$\Rightarrow\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 3 & 3 & 3 \\ x & y & z\end{array}\right|=9(\hat{j}-\hat{k})$
$\Rightarrow \quad(3 z-3 y) \hat{i}+(3 x-3 z) \hat{j}+(3 y-3 x) k=9 \hat{j}-9 k$
$\therefore \quad 3 z-3 y=0 \Rightarrow y=z$ .....(1)
$3 x-3 z=9 \Rightarrow x-z=3$ .....(2)
$3 y-3 x=-9 \Rightarrow y-x=-3$ .....(3)
Clearly $y=z$ is satisfied in
$\therefore \quad \vec{b}=5 \hat{i}+2 \hat{j}+2 \hat{k}$
Hence, the correct answer is option: 'B'.