Question

If $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2\hat{\mathrm{j}}-3\hat{\mathrm{k}}$, $\overrightarrow{\mathrm{b}}=3\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}}$ then the angle between $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$ is:

• (A) $0^{\circ }$
• (B) ${180}^{\circ }$
• (C) ${90}^{\circ }$
• (D) $-{180}^{\circ }$

Answer 1

The Correct Option is C

Explanation:
Given:$\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2\hat{\mathrm{j}}-3\hat{\mathrm{k}}$$\overrightarrow{\mathrm{b}}=3\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}}$So,$(\overrightarrow{a}+\overrightarrow{b})=\hat{i}+2\hat{j}-3\hat{k}+3\hat{i}-\hat{j}+2\hat{k}$$(\overrightarrow{a}+\overrightarrow{b})=(\hat{i}+\widehat{3i})+(\widehat{2j}-\hat{j})+(\widehat{-3k}+\widehat{2k})$$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=4\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$$(\overrightarrow{a}-\overrightarrow{b})=\hat{i}+2\hat{j}-3\hat{k}-3\hat{i}-\hat{j}+2\hat{k}$$(\overrightarrow{a}-\overrightarrow{b})=(\hat{i}-\widehat{3i})+(\widehat{2j}-\widehat{(-j)})+(\widehat{-3k}-\widehat{2k})$$\Rightarrow \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}=-2\hat{\mathrm{i}}+3\hat{\mathrm{j}}-5\hat{\mathrm{k}}$Then,$(\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}-\overrightarrow{b})=(4\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})\cdot (-2\hat{\mathrm{i}}+3\hat{\mathrm{j}}-5\hat{\mathrm{k}})$$(\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}-\overrightarrow{b})=-8+3+5=0$If $(\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}-\overrightarrow{b})=0$Then angle between two vectors is ${90}^{\circ }$.Hence, the correct option is (C).