A) \[\frac{1}{2}\]
B) 1
C) \[\frac{\sqrt{3}}{2}\]
D) \[\frac{\sqrt{3}-1}{2\sqrt{2}}\]
Correct Answer: C
Solution :
\[|\vec{z}+(\vec{z}\times \vec{x}){{|}^{2}}=|\vec{y}{{|}^{2}}\,\,\Rightarrow \,|\vec{z}{{|}^{2}}+|\vec{z}{{|}^{2}}|\vec{x}{{|}^{2}}\,{{\sin }^{2}}\theta =1\] \[\Rightarrow \,|z|=\frac{1}{\sqrt{1+\,{{\sin }^{2}}\theta }}=\frac{2}{\sqrt{7}}\Rightarrow \,\sin \,\theta \,=\frac{\sqrt{3}}{2}\]You need to login to perform this action.
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