Question:
In a trapezium \(ABCD\),
\[
\overrightarrow{BC}=\lambda \overrightarrow{AD}
\]
and
\[
\vec{x}=\overrightarrow{AC}+\overrightarrow{BD}.
\]
If
\[
\vec{x}=p\overrightarrow{AD},
\]
then \(p=\)
In a trapezium \(ABCD\),
\[
\overrightarrow{BC}=\lambda \overrightarrow{AD}
\]
and
\[
\vec{x}=\overrightarrow{AC}+\overrightarrow{BD}.
\]
If
\[
\vec{x}=p\overrightarrow{AD},
\]
then \(p=\)
Show Hint
In vector problems, first express all vectors using a common set of vectors and then simplify by cancellation.
Updated On: Jun 26, 2026
- \(\lambda-1\)
- \(\lambda+1\)
- \(1-\lambda\)
- \(2\lambda-1\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Express \(\overrightarrow{AC}\) using vector addition.
We know that \[ \overrightarrow{AC} = \overrightarrow{AB}+\overrightarrow{BC} \] Given, \[ \overrightarrow{BC}=\lambda\overrightarrow{AD} \] Hence, \[ \overrightarrow{AC} = \overrightarrow{AB}+\lambda\overrightarrow{AD} \]
Step 2: Express \(\overrightarrow{BD}\).
Also, \[ \overrightarrow{BD} = \overrightarrow{BA}+\overrightarrow{AD} \] Since \[ \overrightarrow{BA}=-\overrightarrow{AB}, \] we get \[ \overrightarrow{BD} = -\overrightarrow{AB}+\overrightarrow{AD} \]
Step 3: Add \(\overrightarrow{AC}\) and \(\overrightarrow{BD}\).
Now, \[ \vec{x} = \overrightarrow{AC}+\overrightarrow{BD} \] Substituting the obtained expressions, \[ \vec{x} = (\overrightarrow{AB}+\lambda\overrightarrow{AD}) + (-\overrightarrow{AB}+\overrightarrow{AD}) \]
Step 4: Simplify the expression.
Cancelling \[ \overrightarrow{AB}, \] we get \[ \vec{x} = \lambda\overrightarrow{AD}+\overrightarrow{AD} \] \[ = (\lambda+1)\overrightarrow{AD} \]
Step 5: Compare with the given form.
Given, \[ \vec{x}=p\overrightarrow{AD} \] Comparing, \[ p=\lambda+1 \]
Step 6: Match with the options.
The obtained value is \[ \lambda+1 \]
Step 7: Final conclusion.
Therefore, \[ \boxed{\lambda+1} \]
We know that \[ \overrightarrow{AC} = \overrightarrow{AB}+\overrightarrow{BC} \] Given, \[ \overrightarrow{BC}=\lambda\overrightarrow{AD} \] Hence, \[ \overrightarrow{AC} = \overrightarrow{AB}+\lambda\overrightarrow{AD} \]
Step 2: Express \(\overrightarrow{BD}\).
Also, \[ \overrightarrow{BD} = \overrightarrow{BA}+\overrightarrow{AD} \] Since \[ \overrightarrow{BA}=-\overrightarrow{AB}, \] we get \[ \overrightarrow{BD} = -\overrightarrow{AB}+\overrightarrow{AD} \]
Step 3: Add \(\overrightarrow{AC}\) and \(\overrightarrow{BD}\).
Now, \[ \vec{x} = \overrightarrow{AC}+\overrightarrow{BD} \] Substituting the obtained expressions, \[ \vec{x} = (\overrightarrow{AB}+\lambda\overrightarrow{AD}) + (-\overrightarrow{AB}+\overrightarrow{AD}) \]
Step 4: Simplify the expression.
Cancelling \[ \overrightarrow{AB}, \] we get \[ \vec{x} = \lambda\overrightarrow{AD}+\overrightarrow{AD} \] \[ = (\lambda+1)\overrightarrow{AD} \]
Step 5: Compare with the given form.
Given, \[ \vec{x}=p\overrightarrow{AD} \] Comparing, \[ p=\lambda+1 \]
Step 6: Match with the options.
The obtained value is \[ \lambda+1 \]
Step 7: Final conclusion.
Therefore, \[ \boxed{\lambda+1} \]
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