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

Options

A. `14//sqrt6`
B. `2//sqrt6`
C. `3//sqrt6`
D. `sqrt5`

Explanation

To solve the problem step by step, we will first find the coordinates of the centroid \( G \) of triangle \( BCD \) and then calculate the vector \( \overrightarrow{AG} \) and its length.

### Step 1: Find the coordinates of point \( G \)

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

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

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

Calculating each component:
- \( x \)-coordinate:
\[ \frac{-1 + 5 + 0}{3} = \frac{4}{3} \]
- \( y \)-coordinate:
\[ \frac{4 + 3 - 5}{3} = \frac{2}{3} \]
- \( z \)-coordinate:
\[ \frac{1 + 2 + 4}{3} = \frac{7}{3} \]

Thus, the coordinates of point \( G \) are:
\[ G\left( \frac{4}{3}, \frac{2}{3}, \frac{7}{3} \right) \]

### Step 2: Find the vector \( \overrightarrow{AG} \)

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

The vector \( \overrightarrow{AG} \) is given by:
\[ \overrightarrow{AG} = \overrightarrow{G} - \overrightarrow{A} \]

Substituting the coordinates:
\[ \overrightarrow{AG} = \left( \frac{4}{3}, \frac{2}{3}, \frac{7}{3} \right) - (3, 0, 1) \]

Calculating each component:
- \( x \)-component:
\[ \frac{4}{3} - 3 = \frac{4}{3} - \frac{9}{3} = -\frac{5}{3} \]
- \( y \)-component:
\[ \frac{2}{3} - 0 = \frac{2}{3} \]
- \( z \)-component:
\[ \frac{7}{3} - 1 = \frac{7}{3} - \frac{3}{3} = \frac{4}{3} \]

Thus, the vector \( \overrightarrow{AG} \) is:
\[ \overrightarrow{AG} = \left( -\frac{5}{3}, \frac{2}{3}, \frac{4}{3} \right) \]

### Step 3: Find the length of vector \( \overrightarrow{AG} \)

The length of vector \( \overrightarrow{AG} \) is given by:
\[ |\overrightarrow{AG}| = \sqrt{\left(-\frac{5}{3}\right)^2 + \left(\frac{2}{3}\right)^2 + \left(\frac{4}{3}\right)^2} \]

Calculating each term:
- \( \left(-\frac{5}{3}\right)^2 = \frac{25}{9} \)
- \( \left(\frac{2}{3}\right)^2 = \frac{4}{9} \)
- \( \left(\frac{4}{3}\right)^2 = \frac{16}{9} \)

Adding these:
\[ |\overrightarrow{AG}| = \sqrt{\frac{25}{9} + \frac{4}{9} + \frac{16}{9}} = \sqrt{\frac{45}{9}} = \sqrt{5} \]

### Final Answer
The length of vector \( \overrightarrow{AG} \) is \( \sqrt{5} \).

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