Question
Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ be three vectors such that $\overrightarrow{a} \times \overrightarrow{b} = 2(\overrightarrow{a} \times \overrightarrow{c})$.If $|\overrightarrow{a}| = 1,|\overrightarrow{\text{ }b}| = 4,|\overrightarrow{c}| = 2$, and the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is $60^{\circ}$, then $|\overrightarrow{a}.\overrightarrow{c}|$ is :
Options
Solution
$\overrightarrow{a} \times \overrightarrow{b} - 2(\overrightarrow{a} \times \overrightarrow{c}) = 0$
$$\begin{array}{r} \overrightarrow{a} \times (\overrightarrow{b} - 2\overrightarrow{c}) = 0 \Rightarrow \overrightarrow{\text{ }b} - 2\overrightarrow{c} = \lambda\overrightarrow{a}\#(1) \end{array}$$
$${|\lambda a|^{2} = |\overrightarrow{b} - 2\overrightarrow{c}|^{2} \Rightarrow \lambda^{2}|\overrightarrow{a}|^{2} = b^{2} + 4c^{2} - 4\overrightarrow{\text{ }b} \cdot \overrightarrow{c} }{\lambda^{2} = 16 + 16 - 4.4.2 \cdot \frac{1}{2} }{\lambda^{2} = 16 }{\lambda = \pm 4 }{\because\overrightarrow{b} - 2\overrightarrow{c} = \pm 4\overrightarrow{a} }$$Dot with $\overrightarrow{c} \Rightarrow \overrightarrow{b} \cdot \overrightarrow{c} - 2|\overrightarrow{c}|^{2} = \pm 4(\overrightarrow{a} \cdot \overrightarrow{c})$
4.2. $\frac{1}{2} - 2.4 = \pm 4(\overrightarrow{a}.\overrightarrow{c})$
$$|\overrightarrow{a} \cdot \overrightarrow{c}| = 1$$
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