In a parallelogram OACB,overrightarrow{OA} = | vedclass

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(D) Given that $OACB$ is a parallelogram with $\overrightarrow{OA} = \vec{a}$ and $\overrightarrow{OB} = \vec{b}$.Since $AC$ is parallel to $OB$,the v

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$\frac{\sqrt{15}}{2}$

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(D) Given that $OACB$ is a parallelogram with $\overrightarrow{OA} = \vec{a}$ and $\overrightarrow{OB} = \vec{b}$.Since $AC$ is parallel to $OB$,the vector $\overrightarrow{AC} = \overrightarrow{OB} = \vec{b}$.The position vector of $M$ lies on $AC$. Since $M$ is on $AC$,we can write $\overrightarrow{OM} = \vec{a} + \lambda \vec{b}$ for some scalar $\lambda$.Then $\overrightarrow{BM} = \overrightarrow{OM} - \overrightarrow{OB} = \vec{a} + \lambda \vec{b} - \vec{b} = \vec{a} + (\lambda - 1)\vec{b}$.Since $BM \perp AC$,we have $\overrightarrow{BM} \cdot \overrightarrow{AC} = 0$.$(\vec{a} + (\lambda - 1)\vec{b}) \cdot \vec{b} = 0$.$\vec{a} \cdot \vec{b} + (\lambda - 1)|\vec{b}|^2 = 0$.Given $\vec{a} \cdot \vec{b} = 1$ and $|\vec{b}| = 2$,so $|\vec{b}|^2 = 4$.$1 + (\lambda - 1)(4) = 0 \implies 4(\lambda - 1) = -1 \implies \lambda - 1 = -\frac{1}{4} \implies \lambda = \frac{3}{4}$.Now,$\overrightarrow{BM} = \vec{a} - \frac{1}{4}\vec{b}$.$|\overrightarrow{BM}|^2 = |\vec{a} - \frac{1}{4}\vec{b}|^2 = |\vec{a}|^2 + \frac{1}{16}|\vec{b}|^2 - \frac{2}{4}(\vec{a} \cdot \vec{b})$.$|\overrightarrow{BM}|^2 = 4 + \frac{4}{16} - \frac{1}{2} = 4 + \frac{1}{4} - \frac{1}{2} = \frac{16 + 1 - 2}{4} = \frac{15}{4}$.Therefore,$|\overrightarrow{BM}| = \frac{\sqrt{15}}{2}$.

In a parallelogram $OACB$,$\overrightarrow{OA} = \vec{a}$,$\overrightarrow{OB} = \vec{b}$,and the foot of the perpendicular drawn from point $B$ to $AC$ is $M$. If $\vec{a} \cdot \vec{b} = 1$ and $|\vec{a}| = |\vec{b}| = 2$,then $|\overrightarrow{BM}|$ is:

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