Mathematics - Vector Question with Solution | TestHub

$\begin{array}{rl}& \overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{d}}-\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{d}}=0\\ & (\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})\times \overrightarrow{\mathrm{d}}=0\\ & \overrightarrow{\mathrm{d}}=\mathrm{t}(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})\\ & \overrightarrow{\mathrm{d}}=\mathrm{t}(-2\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}})\\ & |\overrightarrow{\mathrm{d}}|=1\\ & |\mathrm{t}|=\frac{1}{3}\end{array}$
$\begin{array}{rl}& \overrightarrow{\mathrm{c}}\cdot \overrightarrow{\mathrm{a}}=0\\ & \lambda +\mu =0\\ & \mu =-\lambda \\ & \overrightarrow{\mathrm{c}}=\lambda (\hat{\mathrm{j}}-\hat{\mathrm{k}}),|\overrightarrow{\mathrm{c}}{|}^2=2{\lambda }^2\\ & \overrightarrow{\mathrm{c}}\cdot \hat{\mathrm{d}}=1\\ & \mathrm{t}(-2,-1,2)\cdot \lambda (0,1,-1)=1\\ & \lambda \mathrm{t}=\frac{-1}{3}\Rightarrow {\lambda }^2=1\end{array}$
$\begin{array}{rl}& |3\lambda \hat{\mathrm{d}}+\mu \mathrm{c}{|}^2=9{\lambda }^2|\hat{\mathrm{d}}{|}^2+{\mu }^2|\overrightarrow{\mathrm{c}}{|}^2+6\lambda \mu (\hat{\mathrm{d}}\cdot \overrightarrow{\mathrm{c}})\\ & =3{\lambda }^2+2{\lambda }^4\\ & =5\end{array}$