In a trapezium $A B C D, \mathbf{B C}=\lambda \mathbf{A D}$ and $\mathbf{x}=\mathbf{A C}+\mathbf{B D}$. If…
In a trapezium $A B C D, \mathbf{B C}=\lambda \mathbf{A D}$ and $\mathbf{x}=\mathbf{A C}+\mathbf{B D}$. If $\mathbf{x}=p \mathbf{A D}$, then $p=$
- $\lambda-1$
- $\lambda+1$
- $1-\lambda$
- $2 \lambda-1$
Solution
Let position vector of $A, B, C$ and $D$ are a, b, c and d, respectively.
$
\begin{aligned}
\mathbf{B C} & =\lambda \mathbf{A D} \\
\mathbf{c}-\mathbf{b} & =\lambda(\mathbf{d}-\mathbf{a}) \\
\mathbf{x} & =\mathbf{A C}+\mathbf{B D} \\
\mathbf{x} & =\mathbf{c}-\mathbf{a}+\mathbf{d}-\mathbf{b}=\mathbf{c}-\mathbf{b}+\mathbf{d}-\mathbf{a} \\
& =\lambda(\mathbf{d}-\mathbf{a})+\mathbf{l}(\mathbf{d}-\mathbf{a}) [\because from Eq. (i), $c-b=\lambda(d-a)]
\\
& =(\lambda+1)(\mathbf{d}-\mathbf{a})
\end{aligned}
$
On comparing with $\mathbf{x}=p \mathbf{A D}$
$
\begin{aligned}
& \mathbf{x}=p \mathbf{A D} \\
& (\lambda+1)(\mathbf{d}-\mathbf{a})=p(\mathbf{d}-\mathbf{a}) \\
& \Rightarrow \quad p=\lambda+1 \\
&
\end{aligned}
$
Asked in: AP EAMCET 2022 (06 Jul Shift 2)
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