A) \[\frac{\pi }{3}\]
B) \[\frac{\pi }{6}\]
C) \[{{\cos }^{-1}}\frac{1}{\sqrt{3}}\]
D) \[\frac{\pi }{2}\]
Correct Answer: C
Solution :
Since \[\mathbf{a},\,\,\mathbf{b}\] and \[\mathbf{c}\] are mutually perpendicular, so \[\mathbf{a}\,.\,\mathbf{b}=\mathbf{b}\,.\,\mathbf{c}=\mathbf{c}\,.\,\mathbf{a}=0\] Angle between \[\mathbf{a}\] and \[\mathbf{a}+\mathbf{b}+\mathbf{c}\] is \[\cos \theta =\frac{\mathbf{a}.(\mathbf{a}+\mathbf{b}+\mathbf{c})}{|\mathbf{a}||\mathbf{a}+\mathbf{b}+\mathbf{c}|}\] .....(i) Now \[|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=a\] \[|\mathbf{a}+\mathbf{b}+\mathbf{c}{{|}^{2}}={{\mathbf{a}}^{2}}+{{\mathbf{b}}^{2}}+{{\mathbf{c}}^{2}}+2\,\mathbf{a}\,.\,\mathbf{b}+2\,\mathbf{b}\,.\,\mathbf{c}+2\,\mathbf{c}\,.\,\mathbf{a}\] \[={{a}^{2}}+{{a}^{2}}+{{a}^{2}}+0+0+0\] Þ \[|\mathbf{a}+\mathbf{b}+\mathbf{c}{{|}^{2}}=3{{a}^{2}}\Rightarrow |\mathbf{a}+\mathbf{b}+\mathbf{c}|=\sqrt{3}a\] Putting this value in (i), we get \[\theta ={{\cos }^{-1}}\frac{1}{\sqrt{3}}.\]
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