Question

If A = (5, 1, p), B = (1, q, p) and C = (1, -2, 3) are vertices of triangle and G = $(\frac{-4}{3}, \frac{1}{3})$ is its centroid, then the values of p, q, r are respectively

• A. -1, 3, 7/3
• B. -1, -3, 7/3
• C. 1, -3, 7/3
• D. 1, 3, 7/3

Answer 1

Answer: (B)

-1, -3, 7/3

Answer 2

To find the values of p, q, and r, we use the centroid formula for a triangle in 3D space. Given the vertices A = (5, 1, p), B = (1, q, p), and C = (1, -2, 3), the centroid G is calculated as:

$G = \left(\frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3}, \frac{z_A + z_B + z_C}{3}\right)$

Given $G = \left(\frac{-4}{3}, \frac{1}{3}, r\right)$, we can set up the following equations:

1. For the x-coordinate:

$\frac{5 + 1 + 1}{3} = \frac{7}{3}$

This doesn't directly help us find p, q, or r, but it confirms the x-coordinate calculation.

2. For the y-coordinate:

$\frac{1 + q + (-2)}{3} = \frac{1}{3}$

$1 + q - 2 = 1$

$q - 1 = 1$

$q = 2$

3. For the z-coordinate:

$\frac{p + p + 3}{3} = r$

$\frac{2p + 3}{3} = r$

However, there seems to be a mistake in the centroid G. It should be $\left(r, \frac{-4}{3}, \frac{1}{3}\right)$. Let's correct the equations:

1. For the x-coordinate:

$r = \frac{5 + 1 + 1}{3} = \frac{7}{3}$

2. For the y-coordinate:

$\frac{1 + q + (-2)}{3} = \frac{-4}{3}$

$1 + q - 2 = -4$

$q - 1 = -4$

$q = -3$

3. For the z-coordinate:

$\frac{p + p + 3}{3} = \frac{1}{3}$

$2p + 3 = 1$

$2p = -2$

$p = -1$

Thus, $p = -1$, $q = -3$, and $r = \frac{7}{3}$.