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 ( angleGOA) (O being the origin) is equal to :

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 ( angleGOA) (O being the origin) is equal to : (A) (1/√30) (B) (1/√10) (C) (1/√15) (D) (1/ 2 √15)

Tardigrade Admin · Accepted Answer

G is the centroid of Δ ABC G ≡ (2, 4, 2) vecOG = 2 hati + 4 hatj + 2 hatk vecOA = 3 hati - hatk cos ( angle GOA ) = (vecOG . vecOA/| vecOG|| vecOA|) = (1/√15)

Q. 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 :

$\frac{1}{30}$

$\frac{1}{10}$

$\frac{1}{15}$

$\frac{1}{2 15}$

$G$ is the centroid of $Δ A BC$
$G \equiv (2, 4, 2)$
$OG = 2 \hat{i} + 4 \hat{j} + 2 \hat{k}$
$O A = 3 \hat{i} - \hat{k}$
$cos (∠ GO A) = \frac{v ec OG . O A}{∣ OG ∣∣ O A ∣} = \frac{1}{15}$

Q. 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 :

30​1​

10​1​

15​1​

215​1​