Question:
In a \(\triangle ABC\),
\[
|\overrightarrow{CB}|=a,\quad |\overrightarrow{CA}|=b,\quad |\overrightarrow{AB}|=c
\]
and \(CD\) is the median through the vertex \(C\). Then
\[
\overrightarrow{CA}\cdot \overrightarrow{CD}=
\]
In a \(\triangle ABC\),
\[
|\overrightarrow{CB}|=a,\quad |\overrightarrow{CA}|=b,\quad |\overrightarrow{AB}|=c
\]
and \(CD\) is the median through the vertex \(C\). Then
\[
\overrightarrow{CA}\cdot \overrightarrow{CD}=
\]
Show Hint
For a median from a vertex, use the midpoint vector formula:
\[
\overrightarrow{CD}=\frac{1}{2}(\overrightarrow{CA}+\overrightarrow{CB})
\]
Then apply dot product identities.
Updated On: Jun 26, 2026
- \(\dfrac{1}{4}(3a^2+b^2-c^2)\)
- \(\dfrac{1}{4}(a^2+3b^2-c^2)\)
- \(\dfrac{1}{4}(a^2+b^2-3c^2)\)
- \(\dfrac{1}{4}(-3a^2-b^2+c^2)\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Understand the median vector.
Since \(CD\) is the median from vertex \(C\), point \(D\) is the midpoint of \(AB\).
Therefore, \[ \overrightarrow{CD} = \frac{1}{2}\left(\overrightarrow{CA}+\overrightarrow{CB}\right) \]
Step 2: Dot multiply with \(\overrightarrow{CA}\).
We need to find \[ \overrightarrow{CA}\cdot \overrightarrow{CD} \] Substituting the value of \(\overrightarrow{CD}\), \[ \overrightarrow{CA}\cdot \overrightarrow{CD} = \overrightarrow{CA}\cdot \frac{1}{2}\left(\overrightarrow{CA}+\overrightarrow{CB}\right) \] \[ = \frac{1}{2}\left(\overrightarrow{CA}\cdot \overrightarrow{CA} + \overrightarrow{CA}\cdot \overrightarrow{CB}\right) \]
Step 3: Use the magnitude of \(\overrightarrow{CA}\).
Given, \[ |\overrightarrow{CA}|=b \] So, \[ \overrightarrow{CA}\cdot \overrightarrow{CA}=|\overrightarrow{CA}|^2=b^2 \] Thus, \[ \overrightarrow{CA}\cdot \overrightarrow{CD} = \frac{1}{2}\left(b^2+\overrightarrow{CA}\cdot \overrightarrow{CB}\right) \]
Step 4: Find \(\overrightarrow{CA}\cdot \overrightarrow{CB}\).
We know, \[ \overrightarrow{AB} = \overrightarrow{CB}-\overrightarrow{CA} \] Taking magnitudes, \[ |\overrightarrow{AB}|^2 = |\overrightarrow{CB}-\overrightarrow{CA}|^2 \] Since \[ |\overrightarrow{AB}|=c, \] we get \[ c^2=a^2+b^2-2\overrightarrow{CA}\cdot \overrightarrow{CB} \]
Step 5: Solve for \(\overrightarrow{CA}\cdot \overrightarrow{CB}\).
From \[ c^2=a^2+b^2-2\overrightarrow{CA}\cdot \overrightarrow{CB}, \] we get \[ 2\overrightarrow{CA}\cdot \overrightarrow{CB}=a^2+b^2-c^2 \] \[ \overrightarrow{CA}\cdot \overrightarrow{CB} = \frac{a^2+b^2-c^2}{2} \]
Step 6: Substitute in the required expression.
\[ \overrightarrow{CA}\cdot \overrightarrow{CD} = \frac{1}{2} \left[ b^2+\frac{a^2+b^2-c^2}{2} \right] \] \[ = \frac{1}{2} \left[ \frac{2b^2+a^2+b^2-c^2}{2} \right] \] \[ = \frac{a^2+3b^2-c^2}{4} \]
Step 7: Final conclusion.
Therefore, \[ \boxed{\frac{1}{4}(a^2+3b^2-c^2)} \]
Since \(CD\) is the median from vertex \(C\), point \(D\) is the midpoint of \(AB\).
Therefore, \[ \overrightarrow{CD} = \frac{1}{2}\left(\overrightarrow{CA}+\overrightarrow{CB}\right) \]
Step 2: Dot multiply with \(\overrightarrow{CA}\).
We need to find \[ \overrightarrow{CA}\cdot \overrightarrow{CD} \] Substituting the value of \(\overrightarrow{CD}\), \[ \overrightarrow{CA}\cdot \overrightarrow{CD} = \overrightarrow{CA}\cdot \frac{1}{2}\left(\overrightarrow{CA}+\overrightarrow{CB}\right) \] \[ = \frac{1}{2}\left(\overrightarrow{CA}\cdot \overrightarrow{CA} + \overrightarrow{CA}\cdot \overrightarrow{CB}\right) \]
Step 3: Use the magnitude of \(\overrightarrow{CA}\).
Given, \[ |\overrightarrow{CA}|=b \] So, \[ \overrightarrow{CA}\cdot \overrightarrow{CA}=|\overrightarrow{CA}|^2=b^2 \] Thus, \[ \overrightarrow{CA}\cdot \overrightarrow{CD} = \frac{1}{2}\left(b^2+\overrightarrow{CA}\cdot \overrightarrow{CB}\right) \]
Step 4: Find \(\overrightarrow{CA}\cdot \overrightarrow{CB}\).
We know, \[ \overrightarrow{AB} = \overrightarrow{CB}-\overrightarrow{CA} \] Taking magnitudes, \[ |\overrightarrow{AB}|^2 = |\overrightarrow{CB}-\overrightarrow{CA}|^2 \] Since \[ |\overrightarrow{AB}|=c, \] we get \[ c^2=a^2+b^2-2\overrightarrow{CA}\cdot \overrightarrow{CB} \]
Step 5: Solve for \(\overrightarrow{CA}\cdot \overrightarrow{CB}\).
From \[ c^2=a^2+b^2-2\overrightarrow{CA}\cdot \overrightarrow{CB}, \] we get \[ 2\overrightarrow{CA}\cdot \overrightarrow{CB}=a^2+b^2-c^2 \] \[ \overrightarrow{CA}\cdot \overrightarrow{CB} = \frac{a^2+b^2-c^2}{2} \]
Step 6: Substitute in the required expression.
\[ \overrightarrow{CA}\cdot \overrightarrow{CD} = \frac{1}{2} \left[ b^2+\frac{a^2+b^2-c^2}{2} \right] \] \[ = \frac{1}{2} \left[ \frac{2b^2+a^2+b^2-c^2}{2} \right] \] \[ = \frac{a^2+3b^2-c^2}{4} \]
Step 7: Final conclusion.
Therefore, \[ \boxed{\frac{1}{4}(a^2+3b^2-c^2)} \]
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