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Question:

Let A(3, 0, -1), B(2, 10, 6) and C(1, 2, 1) be the vertices of a triangle and M be the midpoint of AC. If G divides BM in the ratio 2:1, then cos(∠GOA) (O being the origin) is equal to:

1√30

16√10

12√15

1√15

Solution:

G is the centroid of ΔABC
G = ( (3+2+1)/3, (0+10+2)/3, (-1+6+1)/3 ) = (2, 4, 2)
→OG = 2î + 4ĵ + 2k̂
→OA = 3î - k̂
cos(∠GOA) = (→OG . →OA) / (|→OG| |→OA|)
= ( (2î + 4ĵ + 2k̂) . (3î - k̂) ) / ( √(2² + 4² + 2²) √(3² + (-1)²) )
= (6 - 2) / (√24 √10)
= 4 / (2√6 √10)
= 2 / (√60)
= 2 / (2√15)
= 1/√15