Consider a triangular pyramid ABCD the position vectors of whone 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 BCT. The length of the vector `bar(AG)` is

Options

A. `(sqrt(17))`
B. `(sqrt(51))/(3)`
C. `(sqrt(51))/(9)`
D. `(sqrt(59))/(4)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the position vectors of points A, B, C, and D.
The position vectors are given as:
- \( A(3, 0, 1) \)
- \( B(-1, 4, 1) \)
- \( C(5, 3, 2) \)
- \( D(0, -5, 4) \)

### Step 2: Find the centroid (G) of triangle BCD.
The centroid \( G \) of triangle formed by points \( B, C, D \) can be calculated using the formula:
\[ G = \left( \frac{x_B + x_C + x_D}{3}, \frac{y_B + y_C + y_D}{3}, \frac{z_B + z_C + z_D}{3} \right) \]

Calculating each coordinate:
- \( x_G = \frac{-1 + 5 + 0}{3} = \frac{4}{3} \)
- \( y_G = \frac{4 + 3 - 5}{3} = \frac{2}{3} \)
- \( z_G = \frac{1 + 2 + 4}{3} = \frac{7}{3} \)

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

### Step 3: Write the position vectors of points A and G.
The position vector of \( A \) is:
\[ A = 3\hat{i} + 0\hat{j} + 1\hat{k} \]
The position vector of \( G \) is:
\[ G = \frac{4}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{7}{3}\hat{k} \]

### Step 4: Find the vector \( \overrightarrow{AG} \).
The vector \( \overrightarrow{AG} \) can be calculated as:
\[ \overrightarrow{AG} = G - A = \left(\frac{4}{3} - 3\right)\hat{i} + \left(\frac{2}{3} - 0\right)\hat{j} + \left(\frac{7}{3} - 1\right)\hat{k} \]

Calculating each component:
- \( \overrightarrow{AG}_x = \frac{4}{3} - 3 = \frac{4}{3} - \frac{9}{3} = -\frac{5}{3} \)
- \( \overrightarrow{AG}_y = \frac{2}{3} \)
- \( \overrightarrow{AG}_z = \frac{7}{3} - 1 = \frac{7}{3} - \frac{3}{3} = \frac{4}{3} \)

Thus, we have:
\[ \overrightarrow{AG} = -\frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{4}{3}\hat{k} \]

### Step 5: 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 the vector \( \overrightarrow{AG} \) is \( \sqrt{5} \).

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