Question

If $\underset{a}{\to },\underset{b}{\to }\text{and}\underset{c}{\to }\text{are unit vectors such that}\underset{a}{\to }.\underset{b}{\to }=\underset{a}{\to }.\underset{c}{\to }=0\text{and the angle between}\underset{b}{\to }\text{and}\underset{c}{\to }\text{is}\frac{\pi }{6},\text{then prove that}$
$\left(i\right)\underset{a}{\to }=\pm 2(\underset{b}{\to }\times \underset{c}{\to })\left(ii\right)[\underset{a}{\to }+\underset{b}{\to }\underset{b}{\to }+\underset{c}{\to }\underset{c}{\to }+\underset{a}{\to }]=\pm 1.$

Answer 1

Given that $\underset{a}{\to }.\underset{b}{\to }=\underset{a}{\to }.\underset{c}{\to }=0$ which implies, $\underset{a}{\to }$ is perpendicular to both $\underset{b}{\to }$ and $\underset{c}{\to }$ (since these three vectors are unit vectors). That is, $\underset{a}{\to }||\underset{b}{\to }\times \underset{c}{\to }\Rightarrow \underset{a}{\to }\lambda (\underset{b}{\to }\times \underset{c}{\to })$
$\therefore |\underset{a}{\to }|=|\lambda ||\underset{b}{\to }\times \underset{c}{\to }|=|\lambda ||\underset{b}{\to }||\underset{c}{\to }|sin\frac{\pi }{6}\Rightarrow 1=|\lambda |\times 1\times 1\times \frac{1}{2}\therefore \lambda =\pm 2\text{Therefore,}\underset{a}{\to }=\pm 2(\underset{b}{\to }\times \underset{c}{\to }).\text{Hence (i) is proved.}$
Also, $[\underset{a}{\to }+\underset{b}{\to }\underset{b}{\to }+\underset{c}{\to }\underset{c}{\to }+\underset{a}{\to }]=\left\{\right(\underset{a}{\to }+\underset{b}{\to })\times (\underset{b}{\to }+\underset{c}{\to }\left)\right\}.(\underset{c}{\to }+\underset{a}{\to })=\{\underset{a}{\to }\times \underset{b}{\to }+\underset{a}{\to }\times \underset{c}{\to }+\underset{b}{\to }\times \underset{b}{\to }+\underset{b}{\to }\times \underset{c}{\to }\}.(\underset{c}{\to }\times \underset{a}{\to })\Rightarrow =\{\underset{a}{\to }\times \underset{b}{\to }+\underset{a}{\to }\times \underset{c}{\to }+\underset{0}{\to }+\underset{b}{\to }\times \underset{c}{\to }\}.(\underset{c}{\to }\times \underset{a}{\to })=\{\underset{a}{\to }\times \underset{b}{\to }.\underset{c}{\to }+\underset{a}{\to }\times \underset{c}{\to }.\underset{c}{\to }+\underset{b}{\to }\times \underset{c}{\to }.\underset{c}{\to }+\underset{a}{\to }\times \underset{b}{\to }.\underset{a}{\to }+\underset{a}{\to }\times \underset{c}{\to }.\underset{a}{\to }+\underset{b}{\to }\times \underset{c}{\to }.\underset{a}{\to }\}\Rightarrow \left[\underset{a}{\to }\underset{b}{\to }\underset{c}{\to }\right]+0+0+0+0+\left[\underset{b}{\to }\underset{c}{\to }\underset{a}{\to }\right]=2\left[\underset{a}{\to }\underset{b}{\to }\underset{c}{\to }\right]=2\{\underset{a}{\to }.(\underset{b}{\to }\times \underset{c}{\to }\left)\right\}\Rightarrow =\left[\underset{a}{\to }\underset{b}{\to }\underset{c}{\to }\right]+0+0+0+0+\left[\underset{b}{\to }\underset{c}{\to }\underset{a}{\to }\right]=2\left[\underset{a}{\to }\underset{b}{\to }\underset{c}{\to }\right]\Rightarrow =2\{\underset{a}{\to }.(\pm \frac{1}{2}\underset{a}{\to })\}=\pm 1|\underset{a}{\to }{|}^2=\pm 1.$
Hence (ii) is proved.