Vectors 4 Question 12
12. Let $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ be three vectors having magnitudes 1 , 1 and 2 , respectively. If $\overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}})+\overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{0}}$, then the actue angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{c}}$ is
(1997, 2M)
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Answer:
Correct Answer: 12. $\frac{\pi}{6}$
Solution:
- Given,
$$ \overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}})+\overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{0}} $$
$$ \begin{array}{ll} \Rightarrow & (\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}}) \overrightarrow{\mathbf{a}}-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}}) \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{0}} \\ \Rightarrow & (2 \cos \theta) \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{0}} \end{array} $$
$$ \begin{array}{lrl} & \Rightarrow & (2 \cos \theta \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{c}})^{2}=(-\overrightarrow{\mathbf{b}})^{2} \\ & \Rightarrow & 4 \cos ^{2} \theta \cdot|\overrightarrow{\mathbf{a}}|^{2}+|\overrightarrow{\mathbf{c}}|^{2}-2 \cdot 2 \cos \theta \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}}=|\overrightarrow{\mathbf{b}}|^{2} \\ \Rightarrow & 4 \cos ^{2} \theta-4-8 \cos ^{2} \theta=1 \\ \Rightarrow & 4 \cos ^{2} \theta=3 \\ \Rightarrow & & \cos \theta= \pm \frac{\sqrt{3}}{2} \end{array} $$
For $\theta$ to be acute,
$$ \cos \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{6} $$
