Question

Prove that (4, 4, 2), (3, 5, 2) and (-1, -1, 2) are vertices of a right angle triangle.

Hint:

{Key Points:}

• For right angle triangle, dot product of two sides = 0
• Alternatively, use Pythagoras theorem: \( a^2 + b^2 = c^2 \)
• All points have same z-coordinate (2), so triangle lies in a plane parallel to XY-plane

Answer 1

Let the points be: \[ A(4, 4, 2), \quad B(3, 5, 2), \quad C(-1, -1, 2) \]
Step 1: Find the vectors representing the sides \[ \vec{AB} = B - A = (3 - 4, 5 - 4, 2 - 2) = (-1, 1, 0) \] \[ \vec{BC} = C - B = (-1 - 3, -1 - 5, 2 - 2) = (-4, -6, 0) \] \[ \vec{CA} = A - C = (4 - (-1), 4 - (-1), 2 - 2) = (5, 5, 0) \]
Step 2: Calculate the dot products For a right angle triangle, one of the dot products of two sides should be zero. Check \( \vec{AB} \cdot \vec{BC} \): \[ \vec{AB} \cdot \vec{BC} = (-1)(-4) + (1)(-6) + (0)(0) = 4 - 6 + 0 = -2 \neq 0 \] Check \( \vec{BC} \cdot \vec{CA} \): \[ \vec{BC} \cdot \vec{CA} = (-4)(5) + (-6)(5) + (0)(0) = -20 - 30 + 0 = -50 \neq 0 \] Check \( \vec{CA} \cdot \vec{AB} \): \[ \vec{CA} \cdot \vec{AB} = (5)(-1) + (5)(1) + (0)(0) = -5 + 5 + 0 = 0 \]
Step 3: Interpret the result Since \( \vec{CA} \cdot \vec{AB} = 0 \), vectors \( \vec{CA} \) and \( \vec{AB} \) are perpendicular to each other. Therefore, angle at A (between CA and AB) is \( 90^\circ \).
Step 4: Conclusion Since one angle is \( 90^\circ \), the triangle formed by points A(4, 4, 2), B(3, 5, 2), and C(-1, -1, 2) is a **right-angled triangle** with the right angle at vertex A. \[ \boxed{\text{The points form a right-angled triangle with right angle at A}} \] Verification using Pythagoras theorem: \[ |AB| = \sqrt{(-1)^2 + 1^2 + 0^2} = \sqrt{1 + 1} = \sqrt{2} \] \[ |AC| = \sqrt{5^2 + 5^2 + 0^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \] \[ |BC| = \sqrt{(-4)^2 + (-6)^2 + 0^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \] Check: \( AB^2 + AC^2 = (\sqrt{2})^2 + (5\sqrt{2})^2 = 2 + 50 = 52 = BC^2 \) Thus, Pythagoras theorem holds, confirming the right angle at A.