Let the three sides of a triangle ABC be given by the vectors 2i^–j^+k^, i^–3j^–5k^ and 3i^–4j^–4k^. Let G be the centroid of the triangle ABC. Then 6(|AG→|2+|BG→|2+|CG→|2) is equal to __________. [2025]
(164)
We have in △ABC, AB→+CA→=BC→
Let PV of A→ be 0→ then AB→=B→–A→
⇒ P.V. of B→=2i^–j^+k^
CA→=A→–C→
P.V. of C→=–i^+3j^+5k^
Now, P.V. of G→=A→+B→+C→3=13(i^+2j^+6k^)
Then AG→=13(i^+2j^+6k^)
⇒ |AG→|2=419
⇒ BG→=(13–2)i^+(23+1)j^+(2–1)k^
⇒ |BG→|2=599
⇒ CG→=(13+1)i^+(23–3)j^+(2–5)k^
⇒ |CG→|2=1469
∴ 6(|AG→|2+|BG→|2+|CG→|2)=6(419+599+1469)=164.