A) \[{{120}^{o}}\]
B) \[{{60}^{o}}\]
C) \[{{90}^{o}}\]
D) \[{{30}^{o}}\]
Correct Answer: C
Solution :
We have, \[|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|\] On squaring both sides, we get \[|\vec{a}+\vec{b}{{|}^{2}}=|\vec{a}-\vec{b}{{|}^{2}}\] \[\Rightarrow \] \[|\vec{a}{{|}^{2}}+|\vec{b}{{|}^{2}}+2\vec{a}.\vec{b}=|\vec{a}{{|}^{2}}+|\vec{b}{{|}^{2}}\] \[-2\vec{a}.\vec{b}\] \[\Rightarrow \] \[4\,\vec{a}.\vec{b}=0\] \[\Rightarrow \] \[\,\vec{a}.\vec{b}=0\] \[\Rightarrow \] \[\,\vec{a}\] and \[\vec{b}\]are perpendicular to each other. So, angle between them is \[{{90}^{o}}\]. Alternative: \[\because \] \[|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|\] \[\therefore \] \[\vec{a}\] and \[\vec{b}\] are perpendicular to each other. So, angle between \[\vec{a}\] and \[\vec{b}\] is \[{{90}^{o}}\].
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