Question

Let $\vec{a}=-\hat{i}-\hat{j}+\hat{k},\vec{a}\cdot \vec{b}=1$ and $\vec{a}\times \vec{b}=\hat{i}-\hat{j}$. Then $\vec{a}-6\vec{b}$ is equal to

• A. $3(\hat{i}-\hat{j}-\hat{k})$
• B. $3(\hat{i}-\hat{j}+\hat{k})$
• C. $3(\hat{i}+\hat{j}-\hat{k})$
• D. $3(\hat{i}+\hat{j}+\hat{k})$

Answer 1

Answer: (D)

$\vec{a}\times \vec{b}=(\hat{i}-\hat{j})$
Taking cross product with $\vec{a}$
$\Rightarrow \vec{a}\times (\vec{a}\times \vec{b})=\vec{a}\times (\hat{i}-\hat{j})$
$\Rightarrow (\vec{a}\cdot \vec{b})\vec{a}-(\vec{a}\cdot \vec{a})\vec{b}=\hat{i}+\hat{j}+2\hat{k}$
$\Rightarrow \vec{a}-3\vec{b}=\hat{i}+\hat{j}+2\hat{k}$
$\Rightarrow 2\vec{a}-6\vec{b}=2\hat{i}+2\hat{j}+4\hat{k}$
$\Rightarrow \vec{a}-6\vec{b}=3\hat{i}+3\hat{j}+3\hat{k}$