Mathematics - Vector Question with Solution | TestHub

$\begin{array}{rl}& \vec{a}=\hat{i}+\hat{j}+\hat{k}\\ & \vec{b}=2\hat{i}+2\hat{j}+\hat{k}\\ & \vec{d}=\vec{a}\times \vec{b}\\ & =-\hat{i}+\hat{j}\\ & |\vec{c}-2\vec{a}{|}^2=8\\ & |\vec{c}{|}^2+4|\vec{a}{|}^2-4\vec{a}\cdot \vec{c}=8\\ & |\vec{c}{|}^2+12-4|\vec{c}|=8\\ & |\vec{c}{|}^2-4|\vec{c}|+4=0\\ & |\vec{c}{|}^2=2\\ & \vec{d}=\vec{a}\times \vec{b}\\ & \vec{d}\times \vec{c}=(\vec{a}\times \vec{b})\times \vec{c}\\ & {\left(|\vec{d}|\times |\vec{c}|\sin \frac{\pi }{4}\right)}^2=((\vec{a}\cdot \vec{c})\vec{b}-(\vec{b}\cdot \vec{c})\vec{a}{)}^2\\ & 4=4|\vec{b}{|}^2+(\vec{b}\cdot \vec{c})2\left(|\vec{a}{|}^2\right)-2(\vec{b}\cdot \vec{c})(\vec{a}\cdot \vec{b})\end{array}$ Let $\vec{b}\cdot \vec{c}=x$ $\begin{array}{rl}& 4=36+3x^2-20x\\ & 3x^2-20x+32=0\\ & x=\frac{8}{3},4\\ & \Rightarrow \vec{b}\cdot \vec{c}=\frac{8}{3},4\\ & \Rightarrow \vec{b}\cdot \vec{c}=\frac{8}{3}\end{array}$ Now, $|10-3\vec{b}\cdot \vec{c}|+|\vec{d}\times \vec{c}{|}^2$ $=|10-8|+(2{)}^2=6$