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

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(D) Given vectors are $\overrightarrow{OA} = \vec{a}$,$\overrightarrow{OB} = \vec{b}$,$\overrightarrow{OC} = 2\vec{a} + 3\vec{b}$,and $\overrightarrow

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$\frac{\pi}{2}$

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(D) Given vectors are $\overrightarrow{OA} = \vec{a}$,$\overrightarrow{OB} = \vec{b}$,$\overrightarrow{OC} = 2\vec{a} + 3\vec{b}$,and $\overrightarrow{OD} = \vec{a} - 2\vec{b}$.First,we find the vectors $\overrightarrow{BD}$ and $\overrightarrow{AC}$:$\overrightarrow{BD} = \overrightarrow{OD} - \overrightarrow{OB} = (\vec{a} - 2\vec{b}) - \vec{b} = \vec{a} - 3\vec{b}$.$\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (2\vec{a} + 3\vec{b}) - \vec{a} = \vec{a} + 3\vec{b}$.Let $	heta$ be the angle between $\overrightarrow{BD}$ and $\overrightarrow{AC}$.Then,$\cos 	heta = \frac{\overrightarrow{BD} \cdot \overrightarrow{AC}}{|\overrightarrow{BD}| |\overrightarrow{AC}|}$.The dot product is $\overrightarrow{BD} \cdot \overrightarrow{AC} = (\vec{a} - 3\vec{b}) \cdot (\vec{a} + 3\vec{b}) = |\vec{a}|^2 - 9|\vec{b}|^2$.Given $|\vec{a}| = 3|\vec{b}|$,we have $|\vec{a}|^2 = 9|\vec{b}|^2$.Substituting this into the dot product: $\overrightarrow{BD} \cdot \overrightarrow{AC} = 9|\vec{b}|^2 - 9|\vec{b}|^2 = 0$.Since the dot product is $0$,$\cos 	heta = 0$,which implies $	heta = \frac{\pi}{2}$.

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

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