Mathematics - Vector Question with Solution | TestHub

$\begin{array}{rl}& \overrightarrow{\mathrm{d}}=\lambda (\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})\\ & \overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{d}}=\lambda (\overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}\cdot \overrightarrow{\mathrm{a}})\\ & 1=\lambda (1+\mathrm{x}+5)\end{array}$ $1=\lambda (x+6)$ ...(1) $\begin{array}{rl}& |\overrightarrow{\mathrm{d}}|=1\\ & |\lambda (\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})|=1\\ & |\lambda ((\mathrm{x}+2)\hat{\mathrm{i}}+6\hat{\mathrm{j}}-2\hat{\mathrm{k}})|=1\\ & {\lambda }^2\left((\mathrm{x}+2{)}^2+6^2+2^2\right)=1\\ & {\mathrm{x}}^2+4\mathrm{x}+4+36+4=(\mathrm{x}+6{)}^2\\ & {\mathrm{x}}^2+4\mathrm{x}+44={\mathrm{x}}^2+12\mathrm{x}+36\\ & 8\mathrm{x}=8,\mathrm{x}=1\\ & \left|\begin{array}{ccc}1& 1& 1\\ 2& 4& -5\\ \mathrm{x}& 2& 3\end{array}\right|=(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})\cdot \overrightarrow{\mathrm{c}}\\ & \left|\begin{array}{ccc}0& 0& 1\\ -2& 9& -4\\ \mathrm{x}-2& -1& 3\end{array}\right|=2-9(\mathrm{x}-2)\\ & =20-9\mathrm{x}\\ & \text{ at }\mathrm{x}=1\\ & 20-9=11\end{array}$