TITLE: Vectors : finding scalar $\mu$ in $\overrightarrow a = \mu \overrightarrow b + 4\overrightarrow c $
QUESTION [1 upvotes]: Non-zero vectors $\overrightarrow a ,\overrightarrow b \& \overrightarrow c $ satisfy $\overrightarrow a .\overrightarrow b  = 0$, $\left( {\overrightarrow b  - \overrightarrow a } \right).\left( {\overrightarrow b  + \overrightarrow c } \right) = 0$ and $2\left| {\overrightarrow b  + \overrightarrow c } \right| = \left| {\overrightarrow b  - \overrightarrow a } \right|$. If $\overrightarrow a  = \mu \overrightarrow b  + 4\overrightarrow c $, then find all possible values of $\mu $.

My approach is as follows
Given $\left( {\overrightarrow b  - \overrightarrow a } \right).\left( {\overrightarrow b  + \overrightarrow c } \right) = 0\& 2\left| {\overrightarrow b  + \overrightarrow c } \right| = \left| {\overrightarrow b  - \overrightarrow a } \right|$
$\overrightarrow u  = \left( {\overrightarrow b  - \overrightarrow a } \right);\overrightarrow v  = \left( {\overrightarrow b  + \overrightarrow c } \right)$
$2\left| {\overrightarrow v } \right| = \left| {\overrightarrow u } \right| \Rightarrow 4\left( {{{\left| {\overrightarrow b } \right|}^2} + {{\left| {\overrightarrow c } \right|}^2} + 2\overrightarrow b .\overrightarrow c } \right) = \left( {{{\left| {\overrightarrow b } \right|}^2} + {{\left| {\overrightarrow a } \right|}^2} - 2\overrightarrow a .\overrightarrow c } \right)$
$\overrightarrow a  = \mu \overrightarrow b  + 4\overrightarrow c  \Rightarrow \overrightarrow b  - \overrightarrow u  = \mu \overrightarrow b  + 4\left( {\overrightarrow v  - \overrightarrow b } \right)$
$ \Rightarrow \overrightarrow b .\overrightarrow v  - \overrightarrow u .\overrightarrow v  = \mu \overrightarrow b .\overrightarrow v  + 4\left( {\overrightarrow v .\overrightarrow v  - \overrightarrow b .\overrightarrow v } \right) \Rightarrow 5\overrightarrow b .\overrightarrow v  = \mu \overrightarrow b .\overrightarrow v  + 4{\left| {\overrightarrow v } \right|^2}$\
$ \Rightarrow 5\overrightarrow b .\left( {\overrightarrow b  + \overrightarrow c } \right) = \mu \overrightarrow b .\left( {\overrightarrow b  + \overrightarrow c } \right) + 4{\left| {\overrightarrow v } \right|^2} \Rightarrow 5\overrightarrow b .\overrightarrow b  + 5\overrightarrow b .\overrightarrow c  = \mu \overrightarrow b .\overrightarrow b  + \mu \overrightarrow b .\overrightarrow c  + 4{\left| {\overrightarrow b } \right|^2} + 4{\left| {\overrightarrow c } \right|^2} + 8\overrightarrow b .\overrightarrow c $
Not able to proceed from here onward

REPLY [2 votes]: Let $\vec a=t \vec i$, $\vec b= s \vec j$ then $\vec c=(t\vec i-\mu s \vec j)/4$.
$$2|\vec b+ \vec c|=|\vec a- \vec b| \implies |s \vec j+(t/4) \vec i-(s\mu/4)\vec j|=|t \vec i- s\vec j|$$
$$\implies (4-\mu)^2s^2+t^2=t^2+s^2\implies (\mu^2-8\mu+12)s^2=3t^2~~~(1)$$
Next $$(\vec b-\vec c).(\vec a- \vec b)=0 \implies  t^2=(4-\mu) s^2~~~(2)$$
From (1) and (2) equate $t^2/s^2$, to get
$$3(4-\mu)=\mu^2-8\mu+12 \implies \mu^2-5\mu=0 \implies \mu =0, 5$$
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