Question

If the segments joining the points $A(a,b)$ and $B(c,d)$ studends an angle at the origin, prove that $\cos \theta =\frac{ac+bd}{\sqrt{(a^2+b^2)(c^2+d^2)}}$ .

Answer 1

$OA=\sqrt{a^2+b^2}$

$OB=\sqrt{c^2+d^2}$

$AB=\sqrt{{(a-c)}^2+{(b-d)}^2}$

$\cos \theta =\frac{O{A}^2+O{B}^2-A{B}^2}{2OA.OB}$

Or, $\cos \theta =\frac{(a^2+b^2)+(c^2+d^2)-{(a-c)}^2-{(b-d)}^2}{2\sqrt{a^2+b^2}\sqrt{c^2+d^2}}$

Or,$\cos \theta =\frac{(a^2+b^2+c^2+d^2)-(a^2+c^2-2ac)+(b^2+d^2-2bd)}{2\sqrt{(a^2+b^2)(c^2+d^2)}}$

Or, $\cos \theta =\frac{2(ac+bd)}{2\sqrt{(a^2+b^2)(c^2+d^2)}}\Rightarrow \cos \theta =\frac{ac+bd}{\sqrt{(a^2+b^2)(c^2+d^2)}}$

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