A) \[\overset{\to }{\mathop{{{\mathbf{a}}^{\mathbf{2}}}}}\,+\overset{\to }{\mathop{{{\mathbf{b}}^{\mathbf{2}}}}}\,-(\overset{\to }{\mathop{\mathbf{a}}}\,\cdot \overset{\to }{\mathop{\mathbf{b}}}\,)\]
B) \[\overset{\to }{\mathop{{{\mathbf{a}}^{\mathbf{2}}}}}\,\overset{\to }{\mathop{{{\mathbf{b}}^{\mathbf{2}}}}}\,-{{(\overset{\to }{\mathop{\mathbf{a}}}\,\cdot \overset{\to }{\mathop{\mathbf{b}}}\,)}^{2}}\]
C) \[\overset{\to }{\mathop{{{\mathbf{a}}^{\mathbf{2}}}}}\,+\overset{\to }{\mathop{{{\mathbf{b}}^{\mathbf{2}}}}}\,-2\overset{\to }{\mathop{\mathbf{a}}}\,\cdot \overset{\to }{\mathop{\mathbf{b}}}\,\]
D) \[\overset{\to }{\mathop{{{\mathbf{a}}^{\mathbf{2}}}}}\,+\overset{\to }{\mathop{{{\mathbf{b}}^{\mathbf{2}}}}}\,-2\overset{\to }{\mathop{\mathbf{a}}}\,\cdot \overset{\to }{\mathop{\mathbf{b}}}\,\]
Correct Answer: B
Solution :
Now, \[{{(\overset{\to }{\mathop{\mathbf{a}}}\,\times \overset{\to }{\mathop{\mathbf{b}}}\,)}^{2}}+{{(\overset{\to }{\mathop{\mathbf{a}}}\,\cdot \overset{\to }{\mathop{\mathbf{b}}}\,)}^{2}}\] \[=\overset{\to }{\mathop{{{\mathbf{a}}^{\mathbf{2}}}}}\,\overset{\to }{\mathop{{{\mathbf{b}}^{\mathbf{2}}}}}\,{{\sin }^{2}}\theta +\overset{\to }{\mathop{{{\mathbf{a}}^{\mathbf{2}}}}}\,\overset{\to }{\mathop{{{\mathbf{b}}^{\mathbf{2}}}}}\,{{\cos }^{2}}\theta \] \[=\overset{\to }{\mathop{{{\mathbf{a}}^{\mathbf{2}}}}}\,\overset{\to }{\mathop{{{\mathbf{b}}^{\mathbf{2}}}}}\,({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )=\overset{\to }{\mathop{{{\mathbf{a}}^{\mathbf{2}}}}}\,\overset{\to }{\mathop{{{\mathbf{b}}^{\mathbf{2}}}}}\,\] \[\Rightarrow \] \[{{(\overset{\to }{\mathop{\mathbf{a}}}\,\times \overset{\to }{\mathop{\mathbf{b}}}\,)}^{2}}=\overset{\to }{\mathop{{{\mathbf{a}}^{\mathbf{2}}}}}\,\overset{\to }{\mathop{{{\mathbf{b}}^{\mathbf{2}}}}}\,-{{(\overset{\to }{\mathop{\mathbf{a}}}\,\cdot \overset{\to }{\mathop{\mathbf{b}}}\,)}^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
