In a trapezium,the vector overrightarrow{BC} | vedclass

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(A) Given that $\overrightarrow{BC} = \lambda \overrightarrow{AD}$.We have $p = \overrightarrow{AC} + \overrightarrow{BD}$.Using vector addition,$\ove

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$\mu = \lambda + 1$

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(A) Given that $\overrightarrow{BC} = \lambda \overrightarrow{AD}$.We have $p = \overrightarrow{AC} + \overrightarrow{BD}$.Using vector addition,$\overrightarrow{BD} = \overrightarrow{BC} + \overrightarrow{CD}$.Substituting this into the expression for $p$:$p = \overrightarrow{AC} + (\overrightarrow{BC} + \overrightarrow{CD}) = (\overrightarrow{AC} + \overrightarrow{CD}) + \overrightarrow{BC}$.Since $\overrightarrow{AC} + \overrightarrow{CD} = \overrightarrow{AD}$,we get:$p = \overrightarrow{AD} + \overrightarrow{BC}$.Substituting $\overrightarrow{BC} = \lambda \overrightarrow{AD}$:$p = \overrightarrow{AD} + \lambda \overrightarrow{AD} = (1 + \lambda) \overrightarrow{AD}$.Given $p = \mu \overrightarrow{AD}$,by comparing the coefficients,we find $\mu = \lambda + 1$.

In a trapezium,the vector $\overrightarrow{BC} = \lambda \overrightarrow{AD}$. We will then find that $p = \overrightarrow{AC} + \overrightarrow{BD}$ is collinear with $\overrightarrow{AD}$. If $p = \mu \overrightarrow{AD}$,then

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