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If $\theta$ is the angle between any two vectors $\vec{a}$ and $\vec{b}$, then $|\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}|=|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|$ when $\theta$ is equal to
(a) 0
(b) $\frac{\pi}{4}$
(c) $\frac{\pi}{2}$
(d) $\pi$
(a) 0
(b) $\frac{\pi}{4}$
(c) $\frac{\pi}{2}$
(d) $\pi$
Solution:
1951 Upvotes
Verified Answer
$\theta$ is the angle between $\vec{a}$ and $\vec{b}$
$\therefore \quad|\vec{a} \cdot \vec{b}|=|\vec{a}||\vec{b}||\cos \theta|$ and
$|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}||\sin \theta|$
we have $|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|$
$\therefore \quad|\vec{a}||\vec{b}||\cos \theta|=|\vec{a}||\vec{b}| \sin \theta$
$\Rightarrow \quad|\cos \theta|=|\sin \theta| \quad$ or $\quad|\tan \theta|=1$
or $\tan \theta=1 \Rightarrow \theta=\frac{\pi}{4}$
Option $(b)$ is correct.
$\therefore \quad|\vec{a} \cdot \vec{b}|=|\vec{a}||\vec{b}||\cos \theta|$ and
$|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}||\sin \theta|$
we have $|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|$
$\therefore \quad|\vec{a}||\vec{b}||\cos \theta|=|\vec{a}||\vec{b}| \sin \theta$
$\Rightarrow \quad|\cos \theta|=|\sin \theta| \quad$ or $\quad|\tan \theta|=1$
or $\tan \theta=1 \Rightarrow \theta=\frac{\pi}{4}$
Option $(b)$ is correct.
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