Vectors 1 Question 6
7. Let $\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}$ be three vectors. A vector $\overrightarrow{\mathbf{v}}$ in the plane of $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, whose projection on $\overrightarrow{\mathbf{c}}$ is $\frac{1}{\sqrt{3}}$, is given by
(2011)
(a) $\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$
(b) $-3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
(c) $3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$
(d) $\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$
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Answer:
Correct Answer: 7. (c)
Solution:
- Let $\overrightarrow{\mathbf{v}}=\overrightarrow{\mathbf{a}}+\lambda \overrightarrow{\mathbf{b}}$
$ \overrightarrow{\mathbf{v}}=(1+\lambda) \hat{\mathbf{i}}+(1-\lambda) \hat{\mathbf{j}}(1+\lambda) \hat{\mathbf{k}} $
Projection of $\overrightarrow{\mathbf{v}}$ on $\overrightarrow{\mathbf{c}}=\frac{1}{\sqrt{3}} \Rightarrow \frac{\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{c}}}{|\overrightarrow{\mathbf{c}}|}=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{(1+\lambda)-(1-\lambda)-(1+\lambda)}{\sqrt{3}}=\frac{1}{\sqrt{3}}$
$\Rightarrow \quad 1+\lambda-1+\lambda-1-\lambda=1$
$ \Rightarrow \quad \lambda-1=1 \Rightarrow \lambda=2 $
$ \therefore \quad \overrightarrow{\mathbf{v}}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}} $
