Mathematics - Vector Question with Solution | TestHub

$\vec{c}=\frac{\vec{a}-\mu \vec{b}}{4}$ and $\vec{a}\cdot \vec{b}=0$

Now, $(\vec{b}-\vec{a})\cdot (\vec{b}+\vec{c})=0\Rightarrow (\vec{b}-\vec{a})\cdot \left(\vec{b}+\frac{\vec{a}-\mu \vec{b}}{4}\right)=0$ $\Rightarrow (4-\mu ){\mathrm{b}}^2={\mathrm{a}}^2(\therefore \mu <4)\dots$ (i)

Again $4|\vec{b}+\vec{c}{|}^2=|\vec{b}-\vec{a}{|}^2\Rightarrow 4{\left|\frac{(4-\mu )\vec{b}+\vec{a}}{4}\right|}^2=|\vec{b}-\vec{a}{|}^2$ $\Rightarrow 4{\left(\frac{4-\mu }{4}\right)}^2{b}^2+\frac{a^2}{4}=b^2+a^2\Rightarrow \left((4-\mu {)}^2-4\right)b^2=3a^2\dots$(ii)

(i) \& (ii) we get $\frac{(4-\mu {)}^2-4}{4-\mu }=\frac{3}{1}\Rightarrow {\mu }^2-5\mu =0$

$\Rightarrow \mu =0$ or 5 but as $\mu <4$, so, $\mu =0$.