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In a trapezoid of the vector $ \vec{BC} = \lambda \vec{AD} $. We will, then find that $ \vec{P} = \vec{AC} + \vec{BD} $ is collinear with $ \vec{AD} $. If $ \vec{P} = \mu \vec{AD} $, then

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In a trapezoid of the vector $\vecBC=λ\vecAD$. We will, then find that $\vecP=\vecAC+\vecBD$ is collinear with $\vecAD$. If $\vecP=μ\vecAD$, then

Updated On: Apr 15, 2026

$\mu=\lambda+1$
$\lambda=\mu+1$
$\lambda+\mu=1$
$\mu=2+\lambda$

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The Correct Option is A

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In a trapezoid of the vector \( \vec{BC} = \lambda \vec{AD} \). We will, then find that \( \vec{P} = \vec{AC} + \vec{BD} \) is collinear with \( \vec{AD} \). If \( \vec{P} = \mu \vec{AD} \), then

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The Correct Option is A

Solution and Explanation

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