Mathematics - Vector Question with Solution | TestHub

$\begin{array}{rl}& \because \vec{a}+2\vec{b}+2\overrightarrow{\mathrm{c}}=\overrightarrow{0}\text{ [Given] }\\ \Rightarrow & \vec{a}+2\overrightarrow{\mathrm{c}}=-2\vec{b}\\ \Rightarrow & (\vec{a}+2\mathrm{c}\overrightarrow{)}\cdot (\vec{a}+2\overrightarrow{\mathrm{c}})=(-2\vec{b})(-2\overrightarrow{\mathrm{b}})\\ \Rightarrow \vec{a}\cdot \vec{a}+4\vec{c}\cdot \vec{c}+4\vec{a}\cdot \vec{c}=4\vec{b}\cdot \vec{b}& \\ \Rightarrow & 1+4+4\vec{a}\cdot \overrightarrow{\mathrm{c}}=4\\ \Rightarrow & \vec{a}\cdot \overrightarrow{\mathrm{c}}=\frac{-1}{4}\\ \because & |\vec{a}\cdot \overrightarrow{\mathrm{c}}{|}^2+|\vec{a}\times \vec{c}{|}^2=1(\vec{a}\text{ is unit vector })\\ & \frac{1}{16}+|\vec{a}\times \vec{c}{|}^2=1\\ & |\vec{a}\times \vec{c}{|}^2=\frac{15}{16}\\ & |\vec{a}\times \vec{c}{|}^2=\frac{\sqrt{15}}{4}\end{array}$