Find out whether the points `(0, 7, 10), (-1,6,6) and (-4,9,6)` are the vertices of a right angled triangle.

Answer

To determine whether the points \( A(0, 7, 10) \), \( B(-1, 6, 6) \), and \( C(-4, 9, 6) \) are the vertices of a right-angled triangle, we will follow these steps:

### Step 1: Calculate the lengths of the sides of the triangle

1. **Length of AB**:
\[ AB = \sqrt{(-1 - 0)^2 + (6 - 7)^2 + (6 - 10)^2} \]
\[ = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} \]
\[ = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2} \]

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

3. **Length of AC**:
\[ AC = \sqrt{(0 - (-4))^2 + (7 - 9)^2 + (10 - 6)^2} \]
\[ = \sqrt{(4)^2 + (-2)^2 + (4)^2} \]
\[ = \sqrt{16 + 4 + 16} = \sqrt{36} = 6 \]

### Step 2: Check the Pythagorean theorem

For the points to form a right-angled triangle, the square of the length of the longest side (hypotenuse) should equal the sum of the squares of the other two sides.

Here, we have:
- \( AB = 3\sqrt{2} \)
- \( BC = 3\sqrt{2} \)
- \( AC = 6 \)

Now, we check if:
\[ AB^2 + BC^2 = AC^2 \]

Calculating each term:
1. \( AB^2 = (3\sqrt{2})^2 = 18 \)
2. \( BC^2 = (3\sqrt{2})^2 = 18 \)
3. \( AC^2 = 6^2 = 36 \)

Now substituting:
\[ AB^2 + BC^2 = 18 + 18 = 36 \]
\[ AC^2 = 36 \]

### Conclusion

Since \( AB^2 + BC^2 = AC^2 \), the points \( A(0, 7, 10) \), \( B(-1, 6, 6) \), and \( C(-4, 9, 6) \) are indeed the vertices of a right-angled triangle, with \( AC \) as the hypotenuse.

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