question_answer

If the line segment joining the points \[A(a,\,b)\] and \[B(c,\,d)\] subtends an angle \[\theta \] at the origin, then \[\cos \theta \] is equal to [IIT 1961]

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 \[AOB,\cos \theta =\frac{{{(OA)}^{2}}+{{(OB)}^{2}}-{{(AB)}^{2}}}{2OA.OB}\] \[=\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}-\{{{(a-c)}^{2}}+{{(b-d)}^{2}}\}}{2\sqrt{{{a}^{2}}+{{b}^{2}}}.\sqrt{{{c}^{2}}+{{d}^{2}}}}\] \[=\frac{ac+bd}{\sqrt{({{a}^{2}}+{{b}^{2}})({{c}^{2}}+{{d}^{2}})}}\].