If $|\mathbf{a}|=1,|\mathbf{b}|=1,|\mathbf{c}|=2$ and $\mathbf{a} \times(\mathbf{a} \times…
If $|\mathbf{a}|=1,|\mathbf{b}|=1,|\mathbf{c}|=2$ and $\mathbf{a} \times(\mathbf{a} \times \mathbf{c})+\mathbf{b}=0$, then $(\mathbf{a} \cdot \mathbf{c})^2=0$
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Solution
If is given that $\mathbf{a} \times(\mathbf{a} \times \mathbf{c})+\mathbf{b}=0$
$
\begin{array}{rlrl}
\Rightarrow & (\mathbf{a} \cdot \mathbf{c}) \mathbf{a}-(\mathbf{a} \cdot \mathbf{a}) \mathbf{c} & =-\mathbf{b} \\
& \text { Let } & \mathbf{a} \cdot \mathbf{c} & =x \\
\Rightarrow & x \mathbf{a}-\mathbf{c} & =-\mathbf{b} \quad[\because|\mathbf{a}|=1 \Rightarrow \mathbf{a} \cdot \mathbf{a}=1]
\end{array}
$
On squaring both sides, we get
$
\begin{aligned}
& x^2|\mathbf{a}|^2+|\mathbf{c}|^2-2 x(\mathbf{a} \cdot \mathbf{c})=|\mathbf{b}|^2 \\
& \left.\Rightarrow \quad x^2+4-2 x^2=1 \quad \text { [as }|\mathbf{b}|=1 \text { and }|\mathbf{c}|=2\right] \\
& \Rightarrow \quad x^2=3 \Rightarrow(\mathbf{a} \cdot \mathbf{c})^2=3
\end{aligned}
$
Hence, option (d) is correct