A) \[\frac{ab+cd}{\sqrt{({{a}^{2}}+{{b}^{2}})\,({{c}^{2}}+{{d}^{2}})}}\]
B) \[\frac{ac+bd}{\sqrt{({{a}^{2}}+{{b}^{2}})\,({{c}^{2}}+{{d}^{2}})}}\]
C) \[\frac{ac-bd}{\sqrt{({{a}^{2}}+{{b}^{2}})\,({{c}^{2}}+{{d}^{2}})}}\]
D) None of these
Correct Answer: B
Solution :
Here \[{{(AB)}^{2}}={{(a-c)}^{2}}+{{(b-d)}^{2}}\] \[{{(OA)}^{2}}={{(a-0)}^{2}}+{{(b-0)}^{2}}={{a}^{2}}+{{b}^{2}}\]and \[{{(OB)}^{2}}={{c}^{2}}+{{d}^{2}}\] Now from triangle \[AOB,\cos \theta =\frac{{{(OA)}^{2}}+{{(OB)}^{2}}-{{(AB)}^{2}}}{2OA.OB}\] \[=\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}-\{{{(a-c)}^{2}}+{{(b-d)}^{2}}\}}{2\sqrt{{{a}^{2}}+{{b}^{2}}}.\sqrt{{{c}^{2}}+{{d}^{2}}}}\] \[=\frac{ac+bd}{\sqrt{({{a}^{2}}+{{b}^{2}})({{c}^{2}}+{{d}^{2}})}}\].You need to login to perform this action.
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