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Let the vectors \( \overrightarrow a\) and \( \overrightarrow b\) be such that \(| \overrightarrow a | = 3\) and \( |\overrightarrow b | = \frac{\large \sqrt 2}{\large 3}\) , then \( \overrightarrow a \times \overrightarrow b\) is a unit vector, if the angle between \( \overrightarrow a\) and \( \overrightarrow b\) is

\[ \begin{array} (A) \frac{\pi}{6} \quad & (B) \frac{\pi}{4} \quad & (C) \frac{\pi}{3} \quad &(D) \frac{\pi}{2} \end{array} \]

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1 Answer

  • $\overrightarrow a\times \overrightarrow b=|\overrightarrow a||\overrightarrow b|\sin\theta\; \hat n$ where $\hat n$ is the unit vector.
Step 1:
Given $|\overrightarrow a |=3$ and $|\overrightarrow b|=\large\frac{\sqrt 2}{3}$ and $\overrightarrow a\times\overrightarrow b$ is a unit vector.
$\overrightarrow a\times \overrightarrow b=|\overrightarrow a||\overrightarrow b|\sin\theta\; \hat n$----(1) where $\hat n$ is the unit vector.
But $\overrightarrow a\times \overrightarrow b$ is also a unit vector.
We know $|\overrightarrow a|=3$ and $|\overrightarrow b|=\large\frac{\sqrt 2}{3}$
Step 2:
Substitute these values in eq(1) we get
$1=3\times\large\frac{\sqrt 2}{3}$$\sin\theta$
$\Rightarrow \sin\theta=\large\frac{1}{\sqrt 2}$
Therefore $\theta=\sin^{-1}\big(\large\frac{1}{\sqrt 2}\big)$
Step 3:
Hence the angle between $\overrightarrow a$ and $\overrightarrow b$ is $\large\frac{\pi}{4}$
Hence B is the correct answer.
answered May 23, 2013 by sreemathi.v

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