Vectors 5 Question 3
3. Let $\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and a unit vector $\overrightarrow{\mathbf{c}}$ be coplanar. If $\overrightarrow{\mathbf{c}}$ is perpendicular to $\overrightarrow{\mathbf{a}}$, then $\overrightarrow{\mathbf{c}}$ is equal to
$(1999,2 M)$
(a) $\frac{1}{\sqrt{2}}(-\hat{\mathbf{j}}+\hat{\mathbf{k}})$
(b) $\frac{1}{\sqrt{3}}(-\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})$
(c) $\frac{1}{\sqrt{5}}(\hat{\mathbf{i}}-2 \hat{\mathbf{j}})$
(d) $\frac{1}{\sqrt{5}}(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})$
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Answer:
Correct Answer: 3. (a)
Solution:
- It is given that $\overrightarrow{\mathbf{c}}$ is coplanar with $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, we take
$$ \overrightarrow{\mathbf{c}}=p \overrightarrow{\mathbf{a}}+q \overrightarrow{\mathbf{b}} $$
where, $p$ and $q$ are scalars.
Since, $\quad \overrightarrow{\mathbf{c}} \perp \overrightarrow{\mathbf{a}} \Rightarrow \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}=0$
Taking dot product of $\overrightarrow{\mathbf{a}}$ in Eq. (i), we get
$$ \begin{aligned} & \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}=p \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}}+q \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}} \Rightarrow 0=p|\overrightarrow{\mathbf{a}}|^{2}+q|\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}}| \\ & {\left[\begin{array}{rl} \because \overrightarrow{\mathbf{a}} & =2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\ |\overrightarrow{\mathbf{a}}| & =\sqrt{2^{2}+1+1}=\sqrt{6} . \\ \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} & =(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ & =2+2-1=3 \end{array}\right] } \\ & \Rightarrow \quad 0=p \cdot 6+q \cdot 3 \quad \Rightarrow \quad q=-2 p \end{aligned} $$
On putting in Eq. (i), we get
$$ \begin{aligned} & \overrightarrow{\mathbf{c}}=p \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}(-2 p) \\ \Rightarrow \quad & \overrightarrow{\mathbf{c}}=p \overrightarrow{\mathbf{a}}-2 p \overrightarrow{\mathbf{b}} \Rightarrow \overrightarrow{\mathbf{c}}=p(\overrightarrow{\mathbf{a}}-2 \overrightarrow{\mathbf{b}}) \\ \Rightarrow \quad & \overrightarrow{\mathbf{c}}=p[(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})-2(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})] \\ \Rightarrow \quad & \overrightarrow{\mathbf{c}}=p(-3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \Rightarrow|\overrightarrow{\mathbf{c}}|=p \sqrt{(-3)^{2}+3^{2}} \\ \Rightarrow \quad & |\overrightarrow{\mathbf{c}}|^{2}=p^{2}(\sqrt{18})^{2} \Rightarrow|\overrightarrow{\mathbf{c}}|^{2}=p^{2} \cdot 18 \\ \Rightarrow \quad & 1=p^{2} \cdot 18 \\ \Rightarrow \quad & \left.p^{2}=\frac{1}{18} \Rightarrow \quad p= \pm \frac{1}{3 \sqrt{2}} \mid=1\right] \\ \therefore \quad & \overrightarrow{\mathbf{c}}= \pm \frac{1(-\hat{\mathbf{j}}+\hat{\mathbf{k}})}{\sqrt{2}} \end{aligned} $$
