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If the segment joining the points \mathrm{A\left ( a,b \right )} and \mathrm{B\left ( c,d \right )} subtend an angle \mathrm{\Theta } at the origin, then \mathrm{cos\ \Theta } \mathrm{ =k/\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}}, where \mathrm{ k=}

Option: 1

\mathrm{ac-bd}


Option: 2

\mathrm{ac+bd}


Option: 3

\mathrm{ad+bc}


Option: 4

\mathrm{ad-bc}


Answers (1)

best_answer

Let \mathrm{O} be the origin. Then \mathrm{OA^2=a^2+b^2, O B^2=c^2+d^2} and \mathrm{A B^2=(c-a)^2+(d-b)^2}
Using cosine formula in \mathrm{\triangle \mathrm{OAB}}, we have \mathrm{\mathrm{AB}^2=\mathrm{OA}^2+\mathrm{OB}^2-2 . \mathrm{OA} \cdot \mathrm{OB} \cos \theta}
\mathrm{ \begin{aligned} & \Rightarrow\left(c-a^2\right)+(d-b)^2=a^2+b^2+c^2+d^2-2 \sqrt{a^2+b^2} \sqrt{c^2+d^2} \cos \theta \\ & \Rightarrow c^2+a^2-2 a c+d^2+b^2-2 b d=a^2+b^2+c^2+d^2-\sqrt{a^2+b^2} \sqrt{c^2+d^2} \cos \theta \\ & \Rightarrow 2(a c+b d)=2 \sqrt{a^2+b^2} \sqrt{c^2+d^2} \cos \theta \\ & \Rightarrow \cos \theta=a c+b d / \sqrt{a^2+b^2} \sqrt{c^2+d^2} \end{aligned} }

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Shailly goel

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