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Question

# Find the angle between the vectors $\stackrel{\to }{a}\mathrm{and}\stackrel{\to }{b}$, where (i) $\stackrel{\to }{a}=\stackrel{^}{i}-\stackrel{^}{j}\mathrm{and}\stackrel{\to }{b}=\stackrel{^}{j}+\stackrel{^}{k}$ (ii) $\stackrel{\to }{a}=3\stackrel{^}{i}-2\stackrel{^}{j}-6\stackrel{^}{k}\mathrm{and}\stackrel{\to }{b}=4\stackrel{^}{i}-\stackrel{^}{j}+8\stackrel{^}{k}$ (iii) $\stackrel{\to }{a}=2\stackrel{^}{i}-\stackrel{^}{j}+2\stackrel{^}{k}\mathrm{and}\stackrel{\to }{b}=4\stackrel{^}{i}+4\stackrel{^}{j}-2\stackrel{^}{k}$ (iv) $\stackrel{\to }{a}=2\stackrel{^}{i}-3\stackrel{^}{j}+\stackrel{^}{k}\mathrm{and}\stackrel{\to }{b}=\stackrel{^}{i}+\stackrel{^}{j}-2\stackrel{^}{k}$ (v) $\stackrel{\to }{a}=\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k},\stackrel{\to }{b}=\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}$

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Solution

## $\left(i\right)\text{Let}\theta \text{be the angle between}\stackrel{\to }{a}\text{and}\stackrel{\to }{b}.\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{a}\right|=\sqrt{{\left(1\right)}^{2}+{\left(-1\right)}^{2}}=\sqrt{2}\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{b}\right|=\sqrt{{\left(1\right)}^{2}+{\left(1\right)}^{2}}=\sqrt{2}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{a}.\stackrel{\to }{b}=0-1+0=-1\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\frac{\stackrel{\to }{a}.\stackrel{\to }{b}}{\left|\stackrel{\to }{a}\right|\left|\stackrel{\to }{b}\right|}=\frac{-1}{\sqrt{2}\sqrt{2}}=\frac{-1}{2}\phantom{\rule{0ex}{0ex}}⇒\theta ={\mathrm{cos}}^{-1}\left(\frac{-1}{2}\right)=\frac{2\mathrm{\pi }}{3}$ $\left(ii\right)\text{Let}\theta \text{be the angle between}\stackrel{\to }{a}\text{and}\stackrel{\to }{b}.\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{a}\right|=\sqrt{{\left(3\right)}^{2}+{\left(-2\right)}^{2}+{\left(-6\right)}^{2}}=\sqrt{49}=7\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{b}\right|=\sqrt{{\left(4\right)}^{2}+{\left(-1\right)}^{2}+{\left(8\right)}^{2}}=\sqrt{81}=9\phantom{\rule{0ex}{0ex}}\stackrel{\to }{a}.\stackrel{\to }{b}=12+2-48=-34\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\frac{\stackrel{\to }{a}.\stackrel{\to }{b}}{\left|\stackrel{\to }{a}\right|\left|\stackrel{\to }{b}\right|}=\frac{-34}{\left(7\right)\left(9\right)}=\frac{-34}{63}\phantom{\rule{0ex}{0ex}}⇒\theta ={\mathrm{cos}}^{-1}\left(\frac{-34}{63}\right)$ $\left(iii\right)\text{Let}\theta \text{be the angle between}\stackrel{\to }{a}\text{and}\stackrel{\to }{b}.\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{a}\right|=\sqrt{{\left(2\right)}^{2}+{\left(-1\right)}^{2}+{\left(2\right)}^{2}}=\sqrt{9}=3\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{b}\right|=\sqrt{{\left(4\right)}^{2}+{\left(4\right)}^{2}+{\left(-2\right)}^{2}}=\sqrt{36}=6\phantom{\rule{0ex}{0ex}}\stackrel{\to }{a}.\stackrel{\to }{b}=8-4-4=0\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\frac{\stackrel{\to }{a}.\stackrel{\to }{b}}{\left|\stackrel{\to }{a}\right|\left|\stackrel{\to }{b}\right|}=\frac{0}{\left(3\right)\left(6\right)}=0\phantom{\rule{0ex}{0ex}}⇒\theta ={\mathrm{cos}}^{-1}\left(0\right)=\frac{\mathrm{\pi }}{2}$ $\left(iv\right)\text{Let}\theta \text{be the angle between}\stackrel{\to }{a}\text{and}\stackrel{\to }{b}.\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{a}\right|=\sqrt{{\left(2\right)}^{2}+{\left(-3\right)}^{2}+{\left(1\right)}^{2}}=\sqrt{14}\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{b}\right|=\sqrt{{\left(1\right)}^{2}+{\left(1\right)}^{2}+{\left(-2\right)}^{2}}=\sqrt{6}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{a}.\stackrel{\to }{b}=2-3-2=-3\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\frac{\stackrel{\to }{a}.\stackrel{\to }{b}}{\left|\stackrel{\to }{a}\right|\left|\stackrel{\to }{b}\right|}=\frac{-3}{\sqrt{14}\sqrt{6}}=\frac{-3}{\sqrt{84}}\phantom{\rule{0ex}{0ex}}⇒\theta ={\mathrm{cos}}^{-1}\left(\frac{-3}{\sqrt{84}}\right)$ $\left(v\right)\text{Let}\theta \text{be the angle between}\stackrel{\to }{a}\text{and}\stackrel{\to }{b}.\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{a}\right|=\sqrt{{\left(1\right)}^{2}+{\left(2\right)}^{2}+{\left(-1\right)}^{2}}=\sqrt{6}\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{b}\right|=\sqrt{{\left(1\right)}^{2}+{\left(-1\right)}^{2}+{\left(1\right)}^{2}}=\sqrt{3}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{a}.\stackrel{\to }{b}=1-2-1=-2\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\frac{\stackrel{\to }{a}.\stackrel{\to }{b}}{\left|\stackrel{\to }{a}\right|\left|\stackrel{\to }{b}\right|}=\frac{-2}{\sqrt{6}\sqrt{3}}=\frac{-2}{\sqrt{18}}=\frac{-\sqrt{2}×\sqrt{2}}{\sqrt{2}×\sqrt{9}}=\frac{-\sqrt{2}}{3}\phantom{\rule{0ex}{0ex}}⇒\theta ={\mathrm{cos}}^{-1}\left(\frac{-\sqrt{2}}{3}\right)$

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