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

We have three vertices $$A(3,0,-1),\;B(2,10,6),\;C(1,2,1).$$ First we find the mid-point $$M$$ of $$AC.$$

Mid-point formula: for endpoints $$P(x_1,y_1,z_1)$$ and $$Q(x_2,y_2,z_2),$$ the mid-point is $$\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2},\dfrac{z_1+z_2}{2}\right).$$

So,

$$M=\left(\dfrac{3+1}{2},\dfrac{0+2}{2},\dfrac{-1+1}{2}\right)=(2,1,0).$$

Next, the segment $$BM$$ is to be divided internally by $$G$$ in the ratio $$BG:GM=2:1.$$

If a point $$P$$ divides $$XY$$ in the ratio $$m:n$$ (measured from $$X$$ towards $$Y$$), then

$$\vec{OP}= \vec{OX}+\dfrac{m}{m+n}\,(\vec{OY}-\vec{OX}).$$

Taking $$X=B,\,Y=M,\,m=2,\,n=1,$$ we get

$$\vec{OG}= \vec{OB}+\dfrac{2}{3}\,(\vec{OM}-\vec{OB}).$$

Compute $$\vec{OM}-\vec{OB}:$$

$$\vec{OM}-\vec{OB}= (2,1,0)-(2,10,6)=(0,-9,-6).$$

Multiplying by $$\dfrac{2}{3}:$$

$$\dfrac{2}{3}(0,-9,-6)=(0,-6,-4).$$

Adding to $$\vec{OB}=(2,10,6):$$

$$\vec{OG}=(2,10,6)+(0,-6,-4)=(2,4,2).$$

The vector $$\vec{OA}$$ is simply $$A-O=(3,0,-1).$$

Now we need $$\cos(\angle GOA),$$ that is the cosine of the angle between $$\vec{OG}$$ and $$\vec{OA}.$$

Dot-product formula: for vectors $$\vec{u},\vec{v},$$

$$\cos\theta=\dfrac{\vec{u}\cdot\vec{v}}{\|\vec{u}\|\;\|\vec{v}\|}.$$

Compute the dot product:

$$\vec{OA}\cdot\vec{OG}= (3)(2)+(0)(4)+(-1)(2)=6-2=4.$$

Compute the magnitudes:

$$\|\vec{OA}\|=\sqrt{3^2+0^2+(-1)^2}=\sqrt{9+1}=\sqrt{10},$$

$$\|\vec{OG}\|=\sqrt{2^2+4^2+2^2}=\sqrt{4+16+4}=\sqrt{24}=2\sqrt6.$$

Substituting in the cosine formula,

$$\cos(\angle GOA)=\dfrac{4}{\sqrt{10}\;(2\sqrt6)}=\dfrac{4}{2\sqrt{60}}=\dfrac{2}{\sqrt{60}}.$$

Simplify $$\sqrt{60}:$$

$$\sqrt{60}=\sqrt{4\cdot15}=2\sqrt{15},$$

so

$$\cos(\angle GOA)=\dfrac{2}{2\sqrt{15}}=\dfrac{1}{\sqrt{15}}.$$

Hence, the correct answer is Option C.