question_answer

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

A) \[\frac{ac+bd}{\sqrt{({{a}^{2}}+{{b}^{2}})({{c}^{2}}+{{d}^{2}})}}\]

B) \[\frac{ab+cd}{\sqrt{({{a}^{2}}+{{b}^{2}})({{c}^{2}}+{{d}^{2}})}}\]

C) \[\frac{ad+bc}{\sqrt{({{a}^{2}}+{{b}^{2}})({{c}^{2}}+{{d}^{2}})}}\]

D) None of these

Correct Answer: A

Solution :

[a] Let the origin be O. so \[O=(0,0)\] Now \[A{{B}^{2}}={{(a-c)}^{2}}+{{(b-d)}^{2}},\] \[O{{A}^{2}}={{(a-0)}^{2}}+{{(b-0)}^{2}}={{a}^{2}}+{{b}^{2}}\] and \[O{{B}^{2}}={{(c-0)}^{2}}+{{(d-0)}^{2}}={{c}^{2}}+{{d}^{2}}\] Now from the \[\Delta AOB:\] \[\cos \theta =\frac{O{{A}^{2}}+O{{B}^{2}}-A{{B}^{2}}}{2OA.OB}\] \[=\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}-\{{{(a-c)}^{2}}+{{(b-d)}^{2}}\}}{2\sqrt{{{a}^{2}}+{{b}^{2}}}\sqrt{{{c}^{2}}+{{d}^{2}}}}\] \[=\frac{2(ac+bd)}{2\sqrt{({{a}^{2}}+{{b}^{2}})({{c}^{2}}+{{d}^{2}})}}=\frac{ac+bd}{\sqrt{({{a}^{2}}+{{b}^{2}})({{c}^{2}}+{{d}^{2}})}}\]