Let the three sides of a triangle ABC be given by the vectors 2 î-ĵ+k̂, î-3 ĵ-5 k̂ and 3 î-4 ĵ-4 k̂. Let G be the centroid of the triangle A B C. Then 6(|AG⃗|²+|BG⃗|²+|CG⃗|²) is equal to __________.

Solution

Maths sir · 4 steps

Do not hunt for A, B and C; the squared side lengths alone fix the distances to the centroid.

1. Check the sides. (2î-ĵ+k̂) + (î-3ĵ-5k̂) = 3î-4ĵ-4k̂, so the three vectors really close into a triangle. We only need their squared lengths.
2. Centroid result. Put G at the origin, so a⃗+b⃗+c⃗ = 0⃗. Squaring gives 2(a⃗·b⃗+b⃗·c⃗+c⃗·a⃗) = -(|a⃗|²+|b⃗|²+|c⃗|²). Expanding |a⃗-b⃗|²+|b⃗-c⃗|²+|c⃗-a⃗|² with this gives three times |a⃗|²+|b⃗|²+|c⃗|².

   |AG⃗|²+|BG⃗|²+|CG⃗|² = 1/3(AB²+BC²+CA²)

3. Side lengths. The squared lengths are 4+1+1 = 6, 1+9+25 = 35 and 9+16+16 = 41. Their sum is 82, so the sum of the squared distances to G is 82/3.
4. So. 6 × 82/3 = 164, so the answer is 164.

The common slip: 492. Forgetting the factor 1/3 and multiplying the side sum directly gives 6 × 82 = 492, not 164.

The idea to keep. For any triangle, GA²+GB²+GC² is one third of the sum of the squared sides. Whenever the centroid appears with squared distances, reach for this result first.

Answer from the NTA final answer key. Solution by PrepWiser.