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Question

# Find the angle between the vectors $\stackrel{\to }{a}=\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\mathrm{and}\stackrel{\to }{b}=\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}.$

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Solution

## $\text{We have}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{a}=\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\mathrm{and}\stackrel{\to }{b}=i+j-k\phantom{\rule{0ex}{0ex}}\text{Let}\theta \text{be the angle between}\stackrel{\to }{a}\text{and}\stackrel{\to }{b}.\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{a}\right|=\sqrt{{\left(1\right)}^{2}+{\left(-1\right)}^{2}+{\left(1\right)}^{2}}=\sqrt{3}\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{b}\right|=\sqrt{{\left(1\right)}^{2}+{\left(1\right)}^{2}+{\left(-1\right)}^{2}}=\sqrt{3}\phantom{\rule{0ex}{0ex}}\mathrm{and}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{a}.\stackrel{\to }{b}=1-1-1=-1\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\frac{\stackrel{\to }{a}.\stackrel{\to }{b}}{\left|\stackrel{\to }{a}\right|\left|\stackrel{\to }{b}\right|}=\frac{-1}{\sqrt{3}\sqrt{3}}=\frac{-1}{3}\phantom{\rule{0ex}{0ex}}\therefore \theta ={\mathrm{cos}}^{-1}\left(\frac{-1}{3}\right)$

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