Vectors 2 Question 20
20. Let $\overrightarrow{\mathbf{A}}, \overrightarrow{\mathbf{B}}$ and $\overrightarrow{\mathbf{C}}$ be unit vectors. If $\overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}}=\overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{C}}=0$ and that the angle between $\overrightarrow{\mathbf{B}}$ and $\overrightarrow{\mathbf{C}}$ is $\pi / 6$.
Then, $\overrightarrow{\mathbf{A}}= \pm 2(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{C}})$.
(1981, 2M)
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Solution:
- Given, $\overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}}=\overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{C}}=0$
$\Rightarrow \overrightarrow{\mathbf{A}}$ is perpendicular to both $\overrightarrow{\mathbf{B}}$ and $\overrightarrow{\mathbf{C}}$.
$\Rightarrow \quad \overrightarrow{\mathbf{A}}=\lambda(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{C}})$
$|\overrightarrow{\mathbf{A}}|=|\lambda||\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{C}}|$, where $\overrightarrow{\mathbf{A}}, \overrightarrow{\mathbf{B}}, \overrightarrow{\mathbf{C}}$ are unit vectors.
$\Rightarrow \quad|\lambda|=\frac{1}{1 \cdot \sin 30^{\circ}} \Rightarrow|\lambda|=2 \Rightarrow \lambda= \pm 2$
$\therefore \quad \overrightarrow{\mathbf{A}}= \pm 2(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{C}})$
Hence, given statement is true.
