Question

Show that the points (-2, 3, 5), (1, 2, 3), and (7, 0, -1) are collinear OR verify that the points (0,7,10), (-1,6,6) and (-4,9,6) are the vertices of a right angled triangle.

Hint:

For points to be collinear, the direction ratios of the vectors formed by the points should be proportional. To check if points form a right angled triangle, verify the Pythagorean theorem.

Answer 1

Step 1: Check if the points are collinear.
For points to be collinear, the vectors formed by them must be parallel. We will find the vectors formed by the points and check if their direction ratios are proportional. Let the points be \( P_1(-2, 3, 5) \), \( P_2(1, 2, 3) \), and \( P_3(7, 0, -1) \). We can find the vectors \( \overrightarrow{P_1P_2} \) and \( \overrightarrow{P_1P_3} \). The vector \( \overrightarrow{P_1P_2} = (1 - (-2), 2 - 3, 3 - 5) = (3, -1, -2) \) The vector \( \overrightarrow{P_1P_3} = (7 - (-2), 0 - 3, -1 - 5) = (9, -3, -6) \) Check if the vectors are proportional: \[ \frac{3}{9} = \frac{-1}{-3} = \frac{-2}{-6} = \frac{1}{3} \] Since the ratios are equal, the points are collinear.
Step 2: Verify if the points are vertices of a right angled triangle.
We are given the points \( P_1(0, 7, 10) \), \( P_2(-1, 6, 6) \), and \( P_3(-4, 9, 6) \). First, we find the squared distances between the points: \[ d_{12}^2 = (0 - (-1))^2 + (7 - 6)^2 + (10 - 6)^2 = 1^2 + 1^2 + 4^2 = 1 + 1 + 16 = 18 \] \[ d_{13}^2 = (0 - (-4))^2 + (7 - 9)^2 + (10 - 6)^2 = 4^2 + (-2)^2 + 4^2 = 16 + 4 + 16 = 36 \] \[ d_{23}^2 = (-1 - (-4))^2 + (6 - 9)^2 + (6 - 6)^2 = 3^2 + (-3)^2 + 0^2 = 9 + 9 + 0 = 18 \] Now, check if the Pythagorean theorem holds: \[ d_{12}^2 + d_{23}^2 = 18 + 18 = 36 = d_{13}^2 \] Since the sum of the squares of two sides equals the square of the third side, the points form a right angled triangle.