Mathematics - Vector Question with Solution | TestHub

$\begin{array}{l}{\overline{\mathrm{r}}}_1=(\lambda \hat{\mathrm{i}}+4\hat{\mathrm{j}}+3\hat{\mathrm{k}})+\alpha (2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}})\\ {\overline{\mathrm{r}}}_2=(2\hat{\mathrm{i}}+4\hat{\mathrm{j}}+7\hat{\mathrm{k}})+\beta (2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}})\end{array}\}\begin{array}{c}\overline{\mathrm{b}}=2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}}\\ {\overline{\mathrm{a}}}_2+\lambda \hat{\mathrm{i}}+4\hat{\mathrm{j}}+3\hat{\mathrm{k}}\\ {\mathrm{a}}_2=2\hat{\mathrm{i}}+4\hat{\mathrm{j}}+7\hat{\mathrm{k}}\end{array}$ Shortest dist. $=\frac{|\overline{\mathrm{b}}\times \left({\overline{\mathrm{a}}}_2-{\overline{\mathrm{a}}}_1\right)|}{|\mathrm{b}|}=\frac{13}{\sqrt{29}}$ $\begin{array}{rl}& |\frac{|(2\hat{i}+3\hat{j}+4\hat{k})\times ((2-\lambda )\hat{i}+4\hat{k})|}{\sqrt{29}}=\frac{13}{\sqrt{29}}\\ & |-8\hat{\mathrm{j}}-3(2-\lambda )\hat{k}+12\hat{i}+4(2-\lambda )\hat{j}|=13\\ & |12\hat{i}-4\lambda \hat{j}+(3\lambda -6)\hat{k}|=13\end{array}$ $\begin{array}{rl}& 144+16{\lambda }^2+(3\lambda -6{)}^2=169\\ & 16{\lambda }^2+(3\lambda -6{)}^2=25=\lambda \Rightarrow =1\end{array}$