Vectors 4 Question 6
6. If $\overrightarrow{\mathbf{a}}=(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}), \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=1$ and $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\hat{\mathbf{j}}-\hat{\mathbf{k}}$, then $\overrightarrow{\mathbf{b}}$ is equal to
(2003, 1M)
(a) $\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$
(b) $2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
(c) $\hat{\mathbf{i}}$
(d) $2 \hat{\mathbf{i}}$
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Answer:
Correct Answer: 6. (c)
Solution:
- We know that, $\overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})=(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}) \overrightarrow{\mathbf{a}}-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}}) \overrightarrow{\mathbf{b}}$
$$ \begin{aligned} & \therefore \quad(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \times(\hat{\mathbf{j}}-\hat{\mathbf{k}})=(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})-(\sqrt{3})^{2} \overrightarrow{\mathbf{b}} \\ & \Rightarrow \quad-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}-3 \overrightarrow{\mathbf{b}} \Rightarrow 3 \overrightarrow{\mathbf{b}}=+3 \hat{\mathbf{i}} \\ & \therefore \quad \overrightarrow{\mathbf{b}}=\hat{\mathbf{i}} \end{aligned} $$
