NEET Physics Mathematical Tools Questions Solved

(C)

$$\begin{gathered} \overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}= \left(2\hat{\mathrm{i}}+3\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)-\left(3\hat{\mathrm{i}}-2\hat{\mathrm{j}}-2\hat{\mathrm{k}}\right) \\ = -\hat{\mathrm{i}}+5\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ \mathrm{Let} \overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}= \overrightarrow{\mathrm{D}} \left(\mathrm{let}\right) \\ \therefore \overrightarrow{\mathrm{D}}= -\hat{\mathrm{i}}+5\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ \left|\overrightarrow{\mathrm{D}}\right|= \sqrt{{\left(-1\right)}^2+{\left(5\right)}^2+{\left(1\right)}^2} = \sqrt{27} \\ \mathrm{To} \mathrm{find} \mathrm{the} \mathrm{angle} \mathrm{between} \overrightarrow{\mathrm{D}} \& \overrightarrow{\mathrm{C}} \\ \left|\overrightarrow{\mathrm{C}}\right|= \sqrt{{\left(\mathrm{p}\right)}^2+{\left(\mathrm{p}\right)}^2+{\left(2\mathrm{p}\right)}^2} = \mathrm{P}\sqrt{6} \\ \overrightarrow{\mathrm{D}}. \overrightarrow{\mathrm{C}}= \mathrm{DC} \mathrm{cos\theta } \\ \left(-\hat{\mathrm{i}}+5\hat{\mathrm{j}}+\hat{\mathrm{k}}\right). \left(\mathrm{p}\hat{\mathrm{i}}+\mathrm{p}\hat{\mathrm{j}}+2\mathrm{p}\hat{\mathrm{k}}\right)= \sqrt{27}\times \mathrm{p}\sqrt{6} \mathrm{cos\theta } \\ -\mathrm{p}+5\mathrm{p}+2\mathrm{p}= \mathrm{p}\sqrt{27\times 6} \mathrm{cos\theta } \\ 6= \sqrt{27\times 6} \mathrm{cos\theta } \\ \mathrm{cos\theta }= \sqrt{\frac{6}{27}} \\ = \sqrt{\frac{2}{9}} \\ \mathrm{cos\theta }= \frac{\sqrt{2}}{3} \mathrm{or} \mathrm{\theta }={\cos }^{-1}\frac{\sqrt{2}}{3} \end{gathered}$$