Straight Line Question 236

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

Options:

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

Show Answer

Answer:

Correct Answer: A

Solution:

  • [a] Let the origin be O. so $ O=(0,0) $ Now $ AB^{2}={{(a-c)}^{2}}+{{(b-d)}^{2}}, $ $ OA^{2}={{(a-0)}^{2}}+{{(b-0)}^{2}}=a^{2}+b^{2} $ and $ OB^{2}={{(c-0)}^{2}}+{{(d-0)}^{2}}=c^{2}+d^{2} $ Now from the $ \Delta 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{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})}} $