Question

If \(A,B,C\) are the vertices of a triangle \(ABC\), \[ AB=2,\qquad BC=3,\qquad CA=4, \] then \[ \overrightarrow{AB}\cdot\overrightarrow{BC} + \overrightarrow{BC}\cdot\overrightarrow{CA} + \overrightarrow{CA}\cdot\overrightarrow{AB} = \]

• A. \(9\)
• B. \(-\dfrac{29}{2}\)
• C. \(-\dfrac{25}{2}\)
• D. \(29\)

Hint:

Whenever you see \[ \overrightarrow{AB}\cdot\overrightarrow{BC} + \overrightarrow{BC}\cdot\overrightarrow{CA} + \overrightarrow{CA}\cdot\overrightarrow{AB}, \] use \[ \overrightarrow{AB} +\overrightarrow{BC} +\overrightarrow{CA} =0 \] and square both sides.

Answer 1

The Correct Option is B

Concept: For any triangle, \[ \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \vec 0. \] Squaring both sides is a powerful technique that generates a relation involving dot products.

Step 1: Use the vector identity of a closed triangle. Since \[ \overrightarrow{AB} +\overrightarrow{BC} +\overrightarrow{CA} = 0, \] taking the square of both sides, \[ (\overrightarrow{AB} +\overrightarrow{BC} +\overrightarrow{CA})^2 = 0. \] Expanding, \[ AB^2+BC^2+CA^2 + 2(\overrightarrow{AB}\cdot\overrightarrow{BC} +\overrightarrow{BC}\cdot\overrightarrow{CA} +\overrightarrow{CA}\cdot\overrightarrow{AB}) =0. \]

Step 2: Substitute the side lengths. Given, \[ AB=2,\qquad BC=3,\qquad CA=4. \] Therefore, \[ AB^2+BC^2+CA^2 = 2^2+3^2+4^2. \] \[ = 4+9+16. \] \[ =29. \] Hence, \[ 29+2S=0, \] where \[ S= \overrightarrow{AB}\cdot\overrightarrow{BC} + \overrightarrow{BC}\cdot\overrightarrow{CA} + \overrightarrow{CA}\cdot\overrightarrow{AB}. \]

Step 3: Solve for \(S\). \[ 2S=-29. \] \[ S=-\frac{29}{2}. \] Conclusion: \[ \boxed{-\frac{29}{2}} \]