Vector Algebra Question 7
Question 7 - 25 January - Shift 2
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
(1) $3(\hat{i}-\hat{j}-\hat{k})$
(2) $3(\hat{i}+\hat{j}+\hat{k})$
(3) $3(\hat{i}-\hat{j}+\hat{k})$
(4) $3(\hat{i}+\hat{j}-\hat{k})$
Show Answer
Answer: (2)
Solution:
Formula: Properties of Scalar product of vectors., Properties of cross product of vectors.
$ \vec{a} \times \vec{b}=(\hat{i}-\hat{j}) $
Taking cross product with $\vec{a}$
$\Rightarrow \vec{a} \times(\vec{a} \times \vec{b})=\vec{a} \times(\hat{i}-\hat{j})$
$\Rightarrow \quad(\vec{a} \cdot \vec{b}) \vec{a}-(\vec{a} \cdot \vec{a}) \vec{b}=\hat{i}+\hat{j}+2 \hat{k}$
$\Rightarrow \vec{a}-3 \vec{b}=\hat{i}+\hat{j}+2 \hat{k}$
$\Rightarrow \quad 2 \vec{a}-6 \vec{b}=2 \hat{i}+2 \hat{j}+4 \hat{k}$
$\Rightarrow \quad \vec{a}-6 \vec{b}=3 \hat{i}+3 \hat{j}+3 \hat{k}$
