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Question

# Evaluate the following: (i) $\left[\stackrel{^}{i}\stackrel{^}{j}\stackrel{^}{k}\right]+\left[\stackrel{^}{j}\stackrel{^}{k}\stackrel{^}{i}\right]+\left[\stackrel{^}{k}\stackrel{^}{i}\stackrel{^}{j}\right]$ (ii) $\left[2\stackrel{^}{i}\stackrel{^}{j}\stackrel{^}{k}\right]+\left[\stackrel{^}{i}\stackrel{^}{k}\stackrel{^}{j}\right]+\left[\stackrel{^}{k}\stackrel{^}{j}2\stackrel{^}{i}\right]$

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Solution

## $\left(\mathrm{i}\right)\mathrm{We}\mathrm{have}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left[\stackrel{^}{i}\stackrel{^}{j}\stackrel{^}{k}\right]+\left[\stackrel{^}{j}\stackrel{^}{k}\stackrel{^}{i}\right]+\left[\stackrel{^}{k}\stackrel{^}{i}\stackrel{^}{j}\right]\phantom{\rule{0ex}{0ex}}=\left[\stackrel{^}{i}\stackrel{^}{j}\stackrel{^}{k}\right]+\left[\stackrel{^}{i}\stackrel{^}{j}\stackrel{^}{k}\right]+\left[\stackrel{^}{i}\stackrel{^}{j}\stackrel{^}{k}\right]\left(\because \left[\stackrel{\to }{a}\stackrel{\to }{b}\stackrel{\to }{c}\right]=\left[\stackrel{\to }{b}\stackrel{\to }{c}\stackrel{\to }{a}\right]=\left[\stackrel{\to }{c}\stackrel{\to }{a}\stackrel{\to }{b}\right]\right)\phantom{\rule{0ex}{0ex}}=3\left[\stackrel{^}{i}\stackrel{^}{j}\stackrel{^}{k}\right]\phantom{\rule{0ex}{0ex}}=3\left(\stackrel{^}{i}×\stackrel{^}{j}\right).\stackrel{^}{k}\left(\because \left[\stackrel{\to }{a}\stackrel{\to }{b}\stackrel{\to }{c}\right]=\left(\stackrel{\to }{a}×\stackrel{\to }{b}\right).\stackrel{\to }{c}\right)\phantom{\rule{0ex}{0ex}}=3\left(\stackrel{^}{k}.\stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}=3\left(1\right)=3$ $\left(\mathrm{ii}\right)\mathrm{We}\mathrm{have}\phantom{\rule{0ex}{0ex}}\left[2\stackrel{^}{i}\stackrel{^}{j}\stackrel{^}{k}\right]+\left[\stackrel{^}{i}\stackrel{^}{k}\stackrel{^}{j}\right]+\left[\stackrel{^}{k}\stackrel{^}{j}2\stackrel{^}{i}\right]\phantom{\rule{0ex}{0ex}}=2\left[\stackrel{^}{i}\stackrel{^}{j}\stackrel{^}{k}\right]+\left[\stackrel{^}{i}\stackrel{^}{k}\stackrel{^}{j}\right]+2\left[\stackrel{^}{k}\stackrel{^}{j}\stackrel{^}{i}\right]\left(\because \left[l\stackrel{\to }{a}m\stackrel{\to }{b}n\stackrel{\to }{c}\right]=lmn\left[\stackrel{\to }{a}\stackrel{\to }{b}\stackrel{\to }{c}\right]\right)\phantom{\rule{0ex}{0ex}}=2\left(\stackrel{^}{i}×\stackrel{^}{j}\right).\stackrel{^}{k}+\left(\stackrel{^}{i}×\stackrel{^}{k}\right).\stackrel{^}{j}+2\left(\stackrel{^}{k}×\stackrel{^}{j}\right).\stackrel{^}{i}\left(\because \left[\stackrel{\to }{a}\stackrel{\to }{b}\stackrel{\to }{c}\right]=\left(\stackrel{\to }{a}×\stackrel{\to }{b}\right).\stackrel{\to }{c}\right)\phantom{\rule{0ex}{0ex}}=2\left(\stackrel{^}{k}.\stackrel{^}{k}\right)+\left(-\stackrel{^}{j}.\stackrel{^}{j}\right)+2\left(-\stackrel{^}{i}.\stackrel{^}{i}\right)\phantom{\rule{0ex}{0ex}}=2\left(1\right)+\left(-1\right)+2\left(-1\right)\phantom{\rule{0ex}{0ex}}=2-1-2=-1$

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