Question

Let \(\vec a,\vec b,\vec c\) be the position vectors of the vertices of a triangle \(ABC\). Through the vertices, lines are drawn parallel to the sides to form the triangle \(A'B'C'\). Then the centroid of \(\triangle A'B'C'\) is

• A. \(\dfrac{\vec a+\vec b+\vec c}{9}\)
• B. \(\dfrac{\vec a+\vec b+\vec c}{6}\)
• C. \(\dfrac{\vec a+\vec b+\vec c}{3}\)
• D. \(\dfrac{2(\vec a+\vec b+\vec c)}{3}\)

Hint:

The centroid of a triangle with position vectors \(\vec a,\vec b,\vec c\) is \(\frac{\vec a+\vec b+\vec c}{3}\). Parallel-side construction preserves the centroid.

Answer 1

The Correct Option is C

Step 1: Understand the construction.
Through each vertex of \(\triangle ABC\), a line is drawn parallel to the opposite side.
These three new lines form a new triangle \(A'B'C'\).

Step 2: Use the standard property.
When a triangle is formed by drawing lines through the vertices of a triangle parallel to the opposite sides, the new triangle has the same centroid as the original triangle.
So, centroid of \(\triangle A'B'C'\) is same as centroid of \(\triangle ABC\).

Step 3: Find the centroid of \(\triangle ABC\).
If the position vectors of the vertices of a triangle are \(\vec a,\vec b,\vec c\), then its centroid is given by
\[ \frac{\vec a+\vec b+\vec c}{3} \]
Therefore, centroid of \(\triangle A'B'C'\) is also
\[ \frac{\vec a+\vec b+\vec c}{3} \]

Step 4: Final conclusion.
Hence,
\[ \boxed{\frac{\vec a+\vec b+\vec c}{3}} \]