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Question

# If θ is the angle between any two vectors $\stackrel{\to }{a}\mathrm{and}\stackrel{\to }{b}$, then $\left|\stackrel{\to }{a}·\stackrel{\to }{b}\right|=\left|\stackrel{\to }{a}×\stackrel{\to }{b}\right|$ when θ is equal to (a) 0 (b) π/4 (c) π/2 (d) π

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Solution

## (b) π/4 $\text{Let}\mathit{\text{θ}}\text{be the angle between}\stackrel{\to }{a}\text{and}\stackrel{\to }{b}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{We know}\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{a}×\stackrel{\to }{b}\right|=\left|\stackrel{\to }{a}\right|\left|\stackrel{\to }{b}\right|\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}\stackrel{\to }{a}.\stackrel{\to }{b}=\left|\stackrel{\to }{a}\right|\left|\stackrel{\to }{b}\right|\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}⇒\left|\stackrel{\to }{a}.\stackrel{\to }{b}\right|=\left|\left|\stackrel{\to }{a}\right|\left|\stackrel{\to }{b}\right|\mathrm{cos}\theta \right|=\left|\stackrel{\text{'}\to }{a}\right|\left|\stackrel{\to }{b}\right|\left|\mathrm{cos}\theta \right|\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{Given:}\left|\stackrel{\to }{a}.\stackrel{\to }{b}\right|=\left|\stackrel{\to }{a}×\stackrel{\to }{b}\right|\phantom{\rule{0ex}{0ex}}⇒\left|\stackrel{\to }{a}\right|\left|\stackrel{\to }{b}\right|\left|\mathrm{cos}\theta \right|=\left|\stackrel{\to }{a}\right|\left|\stackrel{\to }{b}\right|\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}⇒\left|\mathrm{cos}\theta \right|=\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}⇒\theta =\frac{\mathrm{\pi }}{4}$

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