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Step-by-Step Solution
Step 1: Identify the Points and Required Angle
We are given three vertices of a triangle in 3D space:
A(3, 0, –1), B(2, 10, 6), and C(1, 2, 1). O is the origin (0, 0, 0). We want to find
$ \cos\bigl(\angle GOA\bigr) $, where G is the point that divides BM in the ratio 2 : 1 (and hence is the centroid of triangle ABC).
Step 2: Find the Coordinates of G
The centroid G of triangle ABC can be found by
taking the average of the coordinates of A, B, and C:
$
G_x = \frac{3 + 2 + 1}{3} = 2, \quad
G_y = \frac{0 + 10 + 2}{3} = 4, \quad
G_z = \frac{-1 + 6 + 1}{3} = 2.
$
Therefore, $ G = (2,\, 4,\, 2). $
Step 3: Express Vectors OG and OA
Vector $ \overrightarrow{OG} $ is from the origin O to the point G:
$ \overrightarrow{OG} = (2,\, 4,\, 2). $
Vector $ \overrightarrow{OA} $ is from the origin O to the point A:
$ \overrightarrow{OA} = (3,\, 0,\, -1). $
Step 4: Compute Magnitudes of OG and OA
Magnitude of $ \overrightarrow{OG} $:
$
|\overrightarrow{OG}|
= \sqrt{2^2 + 4^2 + 2^2}
= \sqrt{4 + 16 + 4}
= \sqrt{24}.
$
Magnitude of $ \overrightarrow{OA} $:
$
|\overrightarrow{OA}|
= \sqrt{3^2 + 0^2 + (-1)^2}
= \sqrt{9 + 1}
= \sqrt{10}.
$
Step 5: Compute the Dot Product $ \overrightarrow{OG} \cdot \overrightarrow{OA} $
$
\overrightarrow{OG} \cdot \overrightarrow{OA}
= (2)(3) + (4)(0) + (2)(-1)
= 6 + 0 - 2
= 4.
$
Step 6: Calculate $ \cos\bigl(\angle GOA\bigr) $
By definition of the dot product,
$
\overrightarrow{OG} \cdot \overrightarrow{OA}
= |\overrightarrow{OG}|\,
|\overrightarrow{OA}|\,
\cos\bigl(\angle GOA\bigr).
$
So,
$
\cos\bigl(\angle GOA\bigr)
=
\frac{\overrightarrow{OG} \cdot \overrightarrow{OA}}
{|\overrightarrow{OG}|\,
|\overrightarrow{OA}|}
=
\frac{4}{\sqrt{24}\,\sqrt{10}}.
$
Simplify:
$
\frac{4}{\sqrt{24}\,\sqrt{10}}
= \frac{4}{\sqrt{240}}
= \frac{4}{\sqrt{16 \times 15}}
= \frac{4}{4\sqrt{15}}
= \frac{1}{\sqrt{15}}.
$
Step 7: State the Final Answer
Therefore,
$ \cos\bigl(\angle GOA\bigr)
= \frac{1}{\sqrt{15}},
$
which matches the correct choice.
Reference Image
