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Let $\overline{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\overline{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$ be two vectors. If $\overline{\mathrm{c}}$ is a vector such that $\overline{\mathrm{b}} \times \overline{\mathrm{c}}=\overline{\mathrm{b}} \times \overline{\mathrm{a}}$ and $\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}=0$, then $\overline{\mathrm{c}} \cdot \overline{\mathrm{b}}$ is
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Verified Answer
The correct answer is:
$\frac {-1}{2}$
$\begin{aligned}
& \text { Given, } \overline{\mathrm{b}} \times \overline{\mathrm{c}}=\overline{\mathrm{b}} \times \overline{\mathrm{a}} \\
& \Rightarrow \overline{\mathrm{b}} \times(\overline{\mathrm{c}}-\overline{\mathrm{a}})=\overline{0}
\end{aligned}$
$\Rightarrow \overline{\mathrm{b}}$ is parallel to $(\overline{\mathrm{c}}-\overline{\mathrm{a}})$.
$\begin{aligned} & \Rightarrow \overline{\mathrm{c}}-\overline{\mathrm{a}}=\lambda \overline{\mathrm{b}} \text { for some scalar } \lambda \\ & \Rightarrow \overline{\mathrm{c}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}... (i)) \\ & \Rightarrow \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}=\overline{\mathrm{a}} \cdot \overline{\mathrm{a}}+\lambda(\overline{\mathrm{b}} \cdot \overline{\mathrm{a}})\end{aligned}$
$\Rightarrow 0=|\overline{\mathrm{a}}|^2+\lambda(\overline{\mathrm{b}} \cdot \overline{\mathrm{a}}), \quad \cdots[\because \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}=0$ (given) $]$
$\begin{aligned}
& \Rightarrow 0=6 \pm 4 \lambda \\
& \Rightarrow \lambda=-\frac{3}{2}
\end{aligned}$
Substitutifing the value of $\lambda$ in (i), we get
$\begin{aligned}
\bar{c} & =(\hat{i}+2 \hat{j}-\hat{k})-\frac{3}{2}(\hat{i}+\hat{j}-\hat{k}) \\
= & -\frac{1}{2}(\hat{i}-\hat{j}-\hat{k}) \\
\therefore \quad \bar{c} \cdot \bar{b} & =-\frac{1}{2}(\hat{i}-\hat{j}-\hat{k}) \cdot(\hat{i}+\hat{j}-\hat{k}) \\
& =-\frac{1}{2}(1-1+1)=-\frac{1}{2}
\end{aligned}$
& \text { Given, } \overline{\mathrm{b}} \times \overline{\mathrm{c}}=\overline{\mathrm{b}} \times \overline{\mathrm{a}} \\
& \Rightarrow \overline{\mathrm{b}} \times(\overline{\mathrm{c}}-\overline{\mathrm{a}})=\overline{0}
\end{aligned}$
$\Rightarrow \overline{\mathrm{b}}$ is parallel to $(\overline{\mathrm{c}}-\overline{\mathrm{a}})$.
$\begin{aligned} & \Rightarrow \overline{\mathrm{c}}-\overline{\mathrm{a}}=\lambda \overline{\mathrm{b}} \text { for some scalar } \lambda \\ & \Rightarrow \overline{\mathrm{c}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}... (i)) \\ & \Rightarrow \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}=\overline{\mathrm{a}} \cdot \overline{\mathrm{a}}+\lambda(\overline{\mathrm{b}} \cdot \overline{\mathrm{a}})\end{aligned}$
$\Rightarrow 0=|\overline{\mathrm{a}}|^2+\lambda(\overline{\mathrm{b}} \cdot \overline{\mathrm{a}}), \quad \cdots[\because \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}=0$ (given) $]$
$\begin{aligned}
& \Rightarrow 0=6 \pm 4 \lambda \\
& \Rightarrow \lambda=-\frac{3}{2}
\end{aligned}$
Substitutifing the value of $\lambda$ in (i), we get
$\begin{aligned}
\bar{c} & =(\hat{i}+2 \hat{j}-\hat{k})-\frac{3}{2}(\hat{i}+\hat{j}-\hat{k}) \\
= & -\frac{1}{2}(\hat{i}-\hat{j}-\hat{k}) \\
\therefore \quad \bar{c} \cdot \bar{b} & =-\frac{1}{2}(\hat{i}-\hat{j}-\hat{k}) \cdot(\hat{i}+\hat{j}-\hat{k}) \\
& =-\frac{1}{2}(1-1+1)=-\frac{1}{2}
\end{aligned}$
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