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 $\vec{a}$

$\Rightarrow \vec{a} \times(\vec{a} \times \vec{b})=\vec{a} \times(\hat{i}-\hat{j})$

$\Rightarrow \quad(\vec{a} \cdot \vec{b}) \vec{a}-(\vec{a} \cdot \vec{a}) \vec{b}=\hat{i}+\hat{j}+2 \hat{k}$

$\Rightarrow \vec{a}-3 \vec{b}=\hat{i}+\hat{j}+2 \hat{k}$

$\Rightarrow \quad 2 \vec{a}-6 \vec{b}=2 \hat{i}+2 \hat{j}+4 \hat{k}$

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