A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{3}\]
C) \[\frac{\pi }{2}\]
D) \[\frac{3\pi }{2}\]
Correct Answer: B
Solution :
Now,\[2\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{c}}=2(\widehat{\mathbf{i}}+\widehat{\mathbf{j}}-2\widehat{\mathbf{k}})-(-2\widehat{\mathbf{i}}+\widehat{\mathbf{j}}+3\widehat{\mathbf{k}})\] \[=\widehat{\mathbf{j}}+\widehat{\mathbf{k}}\] and \[\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}=-\widehat{\mathbf{i}}+\widehat{\mathbf{j}}+2\widehat{\mathbf{k}}+2\widehat{\mathbf{i}}-\widehat{\mathbf{j}}-\widehat{\mathbf{k}}\] \[=\widehat{\mathbf{i}}+\widehat{\mathbf{k}}\] Let\[\theta \]be the angle between\[2\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{c}}\]and\[\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}\]. \[\therefore \] \[\cos \theta =\frac{(\hat{j}+\hat{k})(\hat{i}+\hat{k})}{\sqrt{{{1}^{2}}+{{1}^{2}}}\sqrt{{{1}^{2}}+{{1}^{2}}}}\] \[\Rightarrow \] \[\cos \theta =\frac{1}{\sqrt{2}\sqrt{2}}=\frac{1}{2}\] \[\Rightarrow \] \[\theta =\frac{\pi }{3}\]
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