question_answer

If the line segment joining the points P (a, b) and Q (c, d) subtends an angle \[\theta \] at the origin, then the value of\[\cos \theta \]is

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

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

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

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

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

Correct Answer: C

Solution :

We know that, \[\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\] \[\therefore \]\[\cos \theta =\frac{\left[ \begin{align}   & {{(\sqrt{{{a}^{2}}+{{b}^{2}}})}^{2}}+{{(\sqrt{{{c}^{2}}+{{d}^{2}}})}^{2}} \\  & -{{(\sqrt{{{(a-c)}^{2}}+{{(b-d)}^{2}}})}^{2}} \\ \end{align} \right]}{2\sqrt{{{a}^{2}}+{{b}^{2}}}\sqrt{{{c}^{2}}+{{d}^{2}}}}\] \[=\frac{\left[ \begin{align}   & {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}-({{a}^{2}}+{{c}^{2}}-2ac \\  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+{{b}^{2}}+{{d}^{2}}-2bd) \\ \end{align} \right]}{2\sqrt{{{a}^{2}}+{{b}^{2}}}\sqrt{{{c}^{2}}+{{d}^{2}}}}\] \[=\frac{ac+bd}{\sqrt{{{a}^{2}}+{{b}^{2}}}\sqrt{{{c}^{2}}+{{d}^{2}}}}\]