A) | u |
B) | u |+| u . a |
C) | u |+| u . b |
D) | u |+ u . (a+b)
Correct Answer: C
Solution :
Let angle between \[\mathbf{a}\] and \[\mathbf{b}\] be \[\theta .\] \[\mathbf{v}=\mathbf{a}\times \mathbf{b}=\,|\mathbf{a}||\mathbf{b}|\sin \theta \,\mathbf{\hat{n}}\] \[\therefore \,\,\,|\mathbf{v}|\,=\sin \theta \], \[\left[ \because \,\,\,|\mathbf{a}|\,=1,\,\,|\mathbf{b}|\,=1,\ \mathbf{\hat{n}}=\frac{(\mathbf{a}\times \mathbf{b})}{|\mathbf{a}\times \mathbf{b}|}=\frac{\mathbf{v}}{|\mathbf{v}|} \right]\] \[\mathbf{u}=\mathbf{a}-(\mathbf{a}.\mathbf{b})\mathbf{b}=\mathbf{a}-\cos \theta \,\mathbf{b}\] \[(\because \,\,\,\mathbf{a}.\mathbf{b}\,=\,|\mathbf{a}||\mathbf{b}|\cos \theta =\cos \theta )\] \[\mathbf{u}\,.\,\mathbf{u}=|\mathbf{u}{{|}^{2}}=1+{{\cos }^{2}}\theta -2\cos \theta \cos \theta ={{\sin }^{2}}\theta \] \[\therefore \,\,\,|\mathbf{u}|=\sin \theta \] \[\mathbf{u}\,.\,\mathbf{a}=\mathbf{a}.\mathbf{a}-\cos \theta \,\mathbf{a}.\mathbf{b}=1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta \] \[\mathbf{u}\,.\,\mathbf{b}=\mathbf{a}.\mathbf{b}-\cos \theta \,\mathbf{b}.\mathbf{b}=\cos \theta -\cos \theta =0\] \[\mathbf{u}\,.\,(\mathbf{a}+\mathbf{b})=(\mathbf{a}-\cos \theta \,\mathbf{b})\,.\,(\mathbf{a}+\mathbf{b})\] \[=1+\cos \theta -{{\cos }^{2}}\theta -\cos \theta \] \[=1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta \] Hence and are correct.
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