In a parallelogram ABCD,|overline{AB}| = a,|o | vedclass

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(C) Let $\vec{AB} = \vec{u}$ and $\vec{AD} = \vec{v}$. Then $|\vec{u}| = a$ and $|\vec{v}| = b$.By the parallelogram law of vector addition,$\vec{AC}

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$\frac{1}{2}(a^2 + b^2 - c^2)$

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(C) Let $\vec{AB} = \vec{u}$ and $\vec{AD} = \vec{v}$. Then $|\vec{u}| = a$ and $|\vec{v}| = b$.By the parallelogram law of vector addition,$\vec{AC} = \vec{AB} + \vec{AD} = \vec{u} + \vec{v}$.Given $|\vec{AC}| = c$,so $|\vec{u} + \vec{v}|^2 = c^2$.Expanding the dot product,we get $|\vec{u}|^2 + |\vec{v}|^2 + 2(\vec{u} \cdot \vec{v}) = c^2$.Substituting the magnitudes,$a^2 + b^2 + 2(\vec{AB} \cdot \vec{AD}) = c^2$.Thus,$\vec{AB} \cdot \vec{AD} = \frac{1}{2}(c^2 - a^2 - b^2)$.We need to find $\overline{DA} \cdot \overline{AB}$.Since $\overline{DA} = -\overline{AD}$,we have $\overline{DA} \cdot \overline{AB} = -(\overline{AD} \cdot \overline{AB}) = -\frac{1}{2}(c^2 - a^2 - b^2) = \frac{1}{2}(a^2 + b^2 - c^2)$.

In a parallelogram $ABCD$,$|\overline{AB}| = a$,$|\overline{AD}| = b$ and $|\overline{AC}| = c$,then the value of $\overline{DA} \cdot \overline{AB}$ is:

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