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Let A(3,0-1) ,B(2,10,6) " and " C(1,2,...

Let `A(3,0-1) ,B(2,10,6) " and " C(1,2,1)` be tge vertices of a trangle and M be the mid-point of AC. If G divide BM in the ratio `2: 1` then cos `(angle GOA) (O " being the origin")` is equal to

A

`(1)/(sqrt(15))`

B

`(1)/(2sqrt(15))`

C

`(1)/(sqrt(30))`

D

`(1)/(6sqrt(10))`

Text Solution

Verified by Experts

The correct Answer is:
A
key idea Use the angle between two non-zero vectors a and b is glven by cos `theta =(a.b)/(|a||b|)` and coordinates of the centroid i.e.
`((x_(1) +x_(2) +x_(3))/(3),(y_(1)+y_(2)+y_(3))/(3),(z_(1)+z_(2)+z_(3))/(3))` of a triangle formed with vertices : `(x_(1),y_(1),z_(1)),(x_(2),y_(2),z_(3)),(x_(3),y_(3),z_(3))`
Given vertices of a `Delta` ABC are A(3,0,-1) B(2,10,6) and C(1,2,1) and a point M is mid-point of A,C .An another point G divides BM in ratio 2:1 so G is the centroid of `Delta `ABC.
`:. G((3+2+1)/(3) , (0+10+2)/(3), (-1+6+1)/(3)) = (2,3,4)`
Now cos `(angle GOA) = (OG.OA)/(|OG||OA|)` where O is the origin .
`:' OG = 2hat(i) + 4hat(j) +2hat(k) rArr |OG| =sqrt(4+16+4) = sqrt(24)`
`" and " |OA| +3hat(i) - hat(k) rArr |OA|=sqrt(9+1)=sqrt(10)`
`" and " OG. OA = 6-2 =4`
`:. cos (angle GOA)= (4)/(sqrt(24)sqrt(10))= (1)/(sqrt(15))`
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