The points O, A, B, C, D are such that overri | vedclass

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(D) We have $\overrightarrow{BD} = \overrightarrow{OD} - \overrightarrow{OB} = (a - 2b) - b = a - 3b$. We have $\overrightarrow{AC} = \overrightarrow{

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None of these

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(D) We have $\overrightarrow{BD} = \overrightarrow{OD} - \overrightarrow{OB} = (a - 2b) - b = a - 3b$. We have $\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (2a + 3b) - a = a + 3b$. Let $	heta$ be the angle between $\overrightarrow{BD}$ and $\overrightarrow{AC}$. Then $\cos 	heta = \frac{\overrightarrow{BD} \cdot \overrightarrow{AC}}{|\overrightarrow{BD}| |\overrightarrow{AC}|} = \frac{(a - 3b) \cdot (a + 3b)}{|\overrightarrow{BD}| |\overrightarrow{AC}|} = \frac{|a|^2 - 9|b|^2}{|\overrightarrow{BD}| |\overrightarrow{AC}|}$. Given $|a| = 3|b|$,we have $|a|^2 = 9|b|^2$. Substituting this into the expression,we get $\cos 	heta = \frac{9|b|^2 - 9|b|^2}{|\overrightarrow{BD}| |\overrightarrow{AC}|} = 0$. Since $\cos 	heta = 0$,the angle $	heta = \frac{\pi}{2}$. Therefore,the correct option is $(d)$.

The points $O, A, B, C, D$ are such that $\overrightarrow{OA} = a, \overrightarrow{OB} = b, \overrightarrow{OC} = 2a + 3b$ and $\overrightarrow{OD} = a - 2b$. If $|a| = 3|b|$,then the angle between $\overrightarrow{BD}$ and $\overrightarrow{AC}$ is

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