Consider a triangular pyramid ABCD the position vectors of whose angular points are `A(3,0,1),B(-1,4,1),C(5,2, 3) and D(0,-5,4)` Let G be the point of intersection of the medians of the triangle BCD. The length of the vec AG is

Options

A. `sqrt17`
B. `sqrt51//3`
C. `3//sqrt6`
D. `sqrt59//4`

Explanation

To find the length of the vector \( \vec{AG} \) where \( G \) is the centroid of triangle \( BCD \), we will follow these steps:

### Step 1: Find the Centroid \( G \) of Triangle \( BCD \)

The coordinates of points \( B \), \( C \), and \( D \) are given as:
- \( B(-1, 4, 1) \)
- \( C(5, 2, 3) \)
- \( D(0, -5, 4) \)

The formula for the centroid \( G \) of a triangle with vertices at \( (x_1, y_1, z_1) \), \( (x_2, y_2, z_2) \), and \( (x_3, y_3, z_3) \) is:
\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right) \]

Substituting the coordinates of points \( B \), \( C \), and \( D \):
\[ G_x = \frac{-1 + 5 + 0}{3} = \frac{4}{3} \]
\[ G_y = \frac{4 + 2 - 5}{3} = \frac{1}{3} \]
\[ G_z = \frac{1 + 3 + 4}{3} = \frac{8}{3} \]

Thus, the coordinates of point \( G \) are:
\[ G\left( \frac{4}{3}, \frac{1}{3}, \frac{8}{3} \right) \]

### Step 2: Find the Position Vector of \( A \)

The position vector of point \( A \) is given as:
\[ A(3, 0, 1) \]

### Step 3: Find the Vector \( \vec{AG} \)

The vector \( \vec{AG} \) is given by:
\[ \vec{AG} = \vec{G} - \vec{A} \]
Substituting the coordinates:
\[ \vec{AG} = \left( \frac{4}{3}, \frac{1}{3}, \frac{8}{3} \right) - (3, 0, 1) \]

Calculating each component:
\[ \vec{AG}_x = \frac{4}{3} - 3 = \frac{4}{3} - \frac{9}{3} = -\frac{5}{3} \]
\[ \vec{AG}_y = \frac{1}{3} - 0 = \frac{1}{3} \]
\[ \vec{AG}_z = \frac{8}{3} - 1 = \frac{8}{3} - \frac{3}{3} = \frac{5}{3} \]

Thus, the vector \( \vec{AG} \) is:
\[ \vec{AG} = \left( -\frac{5}{3}, \frac{1}{3}, \frac{5}{3} \right) \]

### Step 4: Find the Magnitude of \( \vec{AG} \)

The magnitude of a vector \( \vec{v} = (x, y, z) \) is given by:
\[ |\vec{v}| = \sqrt{x^2 + y^2 + z^2} \]

Calculating the magnitude of \( \vec{AG} \):
\[ |\vec{AG}| = \sqrt{\left(-\frac{5}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{5}{3}\right)^2} \]
\[ = \sqrt{\frac{25}{9} + \frac{1}{9} + \frac{25}{9}} = \sqrt{\frac{51}{9}} = \frac{\sqrt{51}}{3} \]

### Final Answer

The length of the vector \( \vec{AG} \) is:
\[ \frac{\sqrt{51}}{3} \]