Let the three sides of triangle ABC be given by the vectors 2i^−j^+k^ (=AB), i^−3j^−5k^ (=BC) and 3i^−4j^−4k^ (=AC). Let G be the centroid of the triangle. Then 6(∣AG∣2+∣BG∣2+∣CG∣2) is equal to \_\_\_\_\_\_\_\_\_\_.
Answer: 164
Solutions · 1
Official solution
Placing A at the origin: A=(0,0,0), B=(2,−1,1), C=(3,−4,−4).
\[G = \frac{A+B+C}3 = \left(\frac53,-\frac53,-1\right)\]
\[|AG|^2 = \frac{25}9+\frac{25}9+1 = \frac{59}9,\quad |BG|^2=\frac19+\frac49+4=\frac{41}9,\quad |CG|^2=\frac{16}9+\frac{49}9+9=\frac{146}9\]
\[6\left(\frac{59+41+146}9\right) = 6\cdot\frac{246}9 = 164\]