Mathematics - Vector Question with Solution | TestHub

$\begin{array}{rl}& \overrightarrow{\mathrm{r}}=\mathrm{k}(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})\\ & \overrightarrow{\mathrm{r}}\cdot \overrightarrow{\mathrm{a}}=3\\ & \overrightarrow{\mathrm{r}}\cdot \overrightarrow{\mathrm{a}}=\mathrm{k}(\overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}\cdot \overrightarrow{\mathrm{a}})\\ & 3=\mathrm{k}(2+6-15+3-2+3\lambda )\end{array}$ $3=\mathrm{k}(-6+3\lambda )$ ...(1) $\overrightarrow{\mathrm{r}}=\mathrm{k}(5\hat{\mathrm{i}}+2\hat{\mathrm{j}}-(5-\lambda )\hat{\mathrm{k}})$ $|\vec{r}|=k\sqrt{25+4+25+{\lambda }^2-10\lambda }=1$ ...(2) $\mathrm{k}=\frac{3}{-6+3\lambda }=\frac{1}{-2+\lambda }$ put in (2) $\begin{array}{rl}& 4+{\lambda }^2-4\lambda =54+{\lambda }^2-10\lambda \\ & 6\lambda =50\\ & 3\lambda =25\end{array}$