If the line segment joining the points A(a, b | vedclass

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(B) Let $O$ be the origin $(0, 0)$. The points are $A(a, b)$ and $B(c, d)$.Using the distance formula,we have:$(OA)^2 = a^2 + b^2$$(OB)^2 = c^2 + d^2$

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

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(B) Let $O$ be the origin $(0, 0)$. The points are $A(a, b)$ and $B(c, d)$.Using the distance formula,we have:$(OA)^2 = a^2 + b^2$$(OB)^2 = c^2 + d^2$$(AB)^2 = (a - c)^2 + (b - d)^2$In $	riangle AOB$,by the Law of Cosines:$\cos 	heta = \frac{(OA)^2 + (OB)^2 - (AB)^2}{2(OA)(OB)}$Substituting the values:$\cos 	heta = \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}}$$\cos 	heta = \frac{a^2 + b^2 + c^2 + d^2 - (a^2 - 2ac + c^2 + b^2 - 2bd + d^2)}{2\sqrt{a^2 + b^2}\sqrt{c^2 + d^2}}$$\cos 	heta = \frac{2ac + 2bd}{2\sqrt{a^2 + b^2}\sqrt{c^2 + d^2}}$$\cos 	heta = \frac{ac + bd}{\sqrt{(a^2 + b^2)(c^2 + d^2)}}$Thus,the correct option is $B$.

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

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