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Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ be unit vectors such that $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{a}.\overrightarrow{c}=0$ and the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is $\Large\frac{\pi}{6}.$ Prove that $\overrightarrow{a}=\pm 2(\overrightarrow{b}\times\overrightarrow{c})$

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  • $ \overrightarrow a.\overrightarrow b=| \overrightarrow a|| \overrightarrow b| \cos \theta$ $ \therefore \cos \theta = \large\frac{ \overrightarrow a. \overrightarrow b}{| \overrightarrow a|| \overrightarrow b|} \Rightarrow \theta = \cos^{-1} \large\frac{ \overrightarrow a. \overrightarrow b}{| \overrightarrow a|| \overrightarrow b|}$
  • If $ \overrightarrow a \perp \overrightarrow b$ then $ \overrightarrow a.\overrightarrow b=0$ and for nonzero vectors if $ \overrightarrow a.\overrightarrow b=0 \Rightarrow \overrightarrow a \perp \overrightarrow b.$
  • If $ \overrightarrow a = a_1\overrightarrow i+a_2\overrightarrow j+a_3\overrightarrow k, \: \overrightarrow b = b_1\overrightarrow i+b_2\overrightarrow j+b_3\overrightarrow k$ then $ \overrightarrow a$ x $ \overrightarrow b = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$
Step 1
$ \overrightarrow a ,\overrightarrow b , \overrightarrow c$ are nonzero vectors.
$ \overrightarrow a.\overrightarrow b = 0 \: and \: \overrightarrow a.\overrightarrow c = 0 . \: \: \: \therefore \overrightarrow a \perp \overrightarrow b \: and \: \overrightarrow a \perp \overrightarrow c$
Step 2
$ \therefore \overrightarrow a$ is perpendicular to the plane containing $ \overrightarrow b \: and \: \overrightarrow c \: or \: \overrightarrow a \parallel \overrightarrow b \times \overrightarrow c$
$ \overrightarrow a, \overrightarrow b, \overrightarrow c$ are unit vectors.
$ \therefore \overrightarrow a = \pm \large\frac{\overrightarrow b \times \overrightarrow c}{|\overrightarrow b \times \overrightarrow c|}$
Step 3
But $ |\overrightarrow a| = |\overrightarrow b| = |\overrightarrow c| = 1$
$ |\overrightarrow b \times \overrightarrow c| = |\overrightarrow b||\overrightarrow c| \sin \large\frac{\pi}{6} = (1)(1).\large\frac{1}{2} = \large\frac{1}{2}$
Step 4
$ \therefore \overrightarrow a = \pm \large\frac{\overrightarrow b \times \overrightarrow c}{\large\frac{1}{2}} = \pm 2(\overrightarrow b \times \overrightarrow c)$
Hence proved
answered Jun 5, 2013 by thanvigandhi_1

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