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 he origin) is equal to:
(A) 1/√30
(B) 1/(2√15)
(C) 1/(6√10)
(D) 1/√15

Answer 1

Option D is correct answer.

Step-by-step explanation:

Given:

A(3, 0, –1), B(2, 10, 6) and C(1, 2, 1) be the vertices of ΔABC.

M is the midpoint of AC.

Also, G divides BM in the ratio 2 : 1.

Let G be the centorid if the triangle.

Coordinates of G = $(\frac{x_1+x_2+x_3}{3} ,\frac{y_1+y_2+y_3}{3} ,\frac{z_1+z_2+z_3}{3} )$

                            = $(\frac{3+2+1}{3} ,\frac{0+10+2}{3} ,\frac{-1+6+1}{3} )$

                            = $(\frac{6}{2} ,\frac{12}{3} ,\frac{6}{3} )$

                            = ( 2, 4, 2)

$cos(\angle GOA)=\frac{\vec{OA} \cdot \vec {OG}}{\vec{|OA|}\cdot \vec{|OG|}}$                 [1]

$\vec{OA}= 3\hat{i}- \hat{k}$

$\vec{|OA|}=\sqrt{3^2+(-1)^2}$

$\vec{|OA|}=\sqrt{9+1}$

$\vec{|OA|}=\sqrt{10}$

$\vec{OG}=2\hat {i}+4\hat{j}+2\hat{k}$

$\vec{OG}=\sqrt{2^2+4^2+2^2}$

$\vec{OG}=\sqrt{4+16+4}$

$\vec{OG}=\sqrt{24}$

$\vec{OA} \cdot \vec{OG}= 6-2 = 4$

Now substituting the values in Eq (1), we get

$cos(\angle GOA)=\frac{4}{\sqrt{10} \cdot \sqrt{24}}$

$cos(\angle GOA)=\frac{\sqrt{4}\sqrt{2}\sqrt{2}}{\sqrt{2}\sqrt{5}\cdot \sqrt{4}\sqrt{2}\sqrt{2}}$                      $[\sqrt{4}\sqrt{2}\sqrt{2}=4]$

$cos(\angle GOA)=\frac{1}{\sqrt{15}}$

Therefore, option D is correct answer.