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If $\mathbf{a}, \mathbf{b}$ and $\overline{\mathbf{c}}$ are three vectors with magnitudes 1,1 and 2 respectively and $\mathbf{a} \times(\mathbf{a} \times \mathbf{c})+\mathbf{b}=0$, then the angle between $a$ and $c$ is
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Verified Answer
The correct answer is:
$\frac{\pi}{6}$
$\begin{aligned}
& \text { }|\mathbf{a}|=1|\mathbf{b}|=1 \quad|\mathbf{c}|=2 \\
& \mathbf{a} \times(\mathbf{a} \times \mathbf{c})+\mathbf{b}=0 \\
& \Rightarrow \quad \mathbf{a} \times(\mathbf{a} \times \mathbf{c})=-\mathbf{b} \\
& \Rightarrow \quad[\mathbf{a} \times(\mathbf{a} \times \mathbf{c})]^2=\mathbf{b}^2 \\
& \Rightarrow \quad[(\mathbf{a} \cdot \mathbf{c}) \mathbf{a}-(\mathbf{a} \cdot \mathbf{a}) \mathbf{c}]^2=\mathbf{b}^2 \\
& \Rightarrow \quad\left[(a c \cos \theta) \mathbf{a}-\mathbf{a}^2 \cdot \mathbf{c}\right]^2=b^2 \\
& \Rightarrow \quad[2 \cdot \cos \theta \cdot \mathbf{a}-\mathbf{c}]^2=b^2 \\
& \Rightarrow 4 \cos ^2 \theta \cdot a^2+c^2-4 \cos \theta \cdot \mathbf{a} \cdot \mathbf{c}=1 \\
& \Rightarrow \quad 4 \cos ^2 \theta+4-4 \cos ^2 \theta \cdot 1 \cdot 2=1 \\
& \Rightarrow \quad-4 \cos ^2 \theta=-3 \\
& \Rightarrow \quad \cos ^2 \theta=\frac{3}{4} \\
& \Rightarrow \quad \cos \theta=\frac{\sqrt{3}}{2}=\frac{\pi}{6} \\
&
\end{aligned}$
& \text { }|\mathbf{a}|=1|\mathbf{b}|=1 \quad|\mathbf{c}|=2 \\
& \mathbf{a} \times(\mathbf{a} \times \mathbf{c})+\mathbf{b}=0 \\
& \Rightarrow \quad \mathbf{a} \times(\mathbf{a} \times \mathbf{c})=-\mathbf{b} \\
& \Rightarrow \quad[\mathbf{a} \times(\mathbf{a} \times \mathbf{c})]^2=\mathbf{b}^2 \\
& \Rightarrow \quad[(\mathbf{a} \cdot \mathbf{c}) \mathbf{a}-(\mathbf{a} \cdot \mathbf{a}) \mathbf{c}]^2=\mathbf{b}^2 \\
& \Rightarrow \quad\left[(a c \cos \theta) \mathbf{a}-\mathbf{a}^2 \cdot \mathbf{c}\right]^2=b^2 \\
& \Rightarrow \quad[2 \cdot \cos \theta \cdot \mathbf{a}-\mathbf{c}]^2=b^2 \\
& \Rightarrow 4 \cos ^2 \theta \cdot a^2+c^2-4 \cos \theta \cdot \mathbf{a} \cdot \mathbf{c}=1 \\
& \Rightarrow \quad 4 \cos ^2 \theta+4-4 \cos ^2 \theta \cdot 1 \cdot 2=1 \\
& \Rightarrow \quad-4 \cos ^2 \theta=-3 \\
& \Rightarrow \quad \cos ^2 \theta=\frac{3}{4} \\
& \Rightarrow \quad \cos \theta=\frac{\sqrt{3}}{2}=\frac{\pi}{6} \\
&
\end{aligned}$
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