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}$
$\Rightarrow \vec{a}-6 \vec{b}=3 \hat{i}+3 \hat{j}+3 \hat{k}$