(a) show that the triangle with vetices (0,7,10) (-1,6,6) and (-4,9,6) is right angled (b) are the points A(3,6,9) B (10,20,30) and C(25,41,5) the vertices of a right angled triangle

Answer

To solve the given problem, we will break it down into two parts as specified in the question.

### Part (a): Show that the triangle with vertices \( A(0, 7, 10) \), \( B(-1, 6, 6) \), and \( C(-4, 9, 6) \) is right-angled.

1. **Calculate the distances between the points:**
- Distance \( AB \):
\[ AB = \sqrt{(-1 - 0)^2 + (6 - 7)^2 + (6 - 10)^2} \]
\[ = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} = \sqrt{1 + 1 + 16} = \sqrt{18} \]

- Distance \( AC \):
\[ AC = \sqrt{(-4 - 0)^2 + (9 - 7)^2 + (6 - 10)^2} \]
\[ = \sqrt{(-4)^2 + (2)^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} \]

- Distance \( BC \):
\[ BC = \sqrt{(-4 - (-1))^2 + (9 - 6)^2 + (6 - 6)^2} \]
\[ = \sqrt{(-3)^2 + (3)^2 + (0)^2} = \sqrt{9 + 9 + 0} = \sqrt{18} \]

2. **Check if the triangle is right-angled using Pythagoras theorem:**
- We need to check if \( AB^2 + BC^2 = AC^2 \):
\[ AB^2 = 18, \quad BC^2 = 18, \quad AC^2 = 36 \]
\[ AB^2 + BC^2 = 18 + 18 = 36 = AC^2 \]

3. **Conclusion:**
- Since \( AB^2 + BC^2 = AC^2 \), the triangle \( ABC \) is right-angled.

### Part (b): Are the points \( A(3, 6, 9) \), \( B(10, 20, 30) \), and \( C(25, 41, 5) \) the vertices of a right-angled triangle?

1. **Calculate the distances between the points:**
- Distance \( AB \):
\[ AB = \sqrt{(10 - 3)^2 + (20 - 6)^2 + (30 - 9)^2} \]
\[ = \sqrt{(7)^2 + (14)^2 + (21)^2} = \sqrt{49 + 196 + 441} = \sqrt{686} \]

- Distance \( AC \):
\[ AC = \sqrt{(25 - 3)^2 + (41 - 6)^2 + (5 - 9)^2} \]
\[ = \sqrt{(22)^2 + (35)^2 + (-4)^2} = \sqrt{484 + 1225 + 16} = \sqrt{1725} \]

- Distance \( BC \):
\[ BC = \sqrt{(25 - 10)^2 + (41 - 20)^2 + (5 - 30)^2} \]
\[ = \sqrt{(15)^2 + (21)^2 + (-25)^2} = \sqrt{225 + 441 + 625} = \sqrt{1291} \]

2. **Check if the triangle is right-angled using Pythagoras theorem:**
- We need to check if \( AB^2 + BC^2 = AC^2 \):
\[ AB^2 = 686, \quad BC^2 = 1291, \quad AC^2 = 1725 \]
\[ AB^2 + BC^2 = 686 + 1291 = 1977 \neq 1725 \]

3. **Conclusion:**
- Since \( AB^2 + BC^2 \neq AC^2 \), the triangle \( ABC \) is not right-angled.