Mathematics - Vector Question with Solution | TestHub

Let's assume $\begin{array}{rl}\vec{v}& =\vec{a}+\vec{b}+\hat{i}\\ & =5\hat{i}+3\hat{j}+\hat{k}\end{array}$ and $\overrightarrow{\mathrm{c}}+\hat{\mathrm{i}}=\overrightarrow{\mathrm{p}}$ So, $\begin{array}{rl}& \overrightarrow{\mathrm{p}}\times \overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{p}}\\ & \overrightarrow{\mathrm{p}}\times \overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{p}}\times \overrightarrow{\mathrm{a}}=\overrightarrow{0}\\ & \overrightarrow{\mathrm{p}}\times (\overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{a}})=\overrightarrow{0}\\ & \Rightarrow \overrightarrow{\mathrm{p}}=\lambda (\overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{a}})\\ & \overrightarrow{\mathrm{c}}+\mathrm{i}=\lambda (7\hat{\mathrm{i}}+8\hat{\mathrm{j}})\\ & \overline{\mathrm{a}}\cdot \overline{\mathrm{c}}+\overline{\mathrm{a}}\cdot \hat{\mathrm{i}}=\lambda \overline{\mathrm{a}}\cdot (7\hat{\mathrm{i}}+8\hat{\mathrm{j}})\\ & -29+2=\lambda (14+40)\\ & \lambda =-\frac{1}{2}\end{array}$ $\begin{array}{c}\overrightarrow{\mathrm{c}}\cdot (-2\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\hat{\mathrm{i}}\cdot (-2\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=\lambda (7\hat{\mathrm{i}}+8\hat{\mathrm{j}})\cdot (-2\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\\ =-\frac{1}{2}(-14+8)+2=5\end{array}$