Question

If \(A(-1,3,2), B(2,3,5), C(3,5,-2)\) are vertices of a triangle ABC, then angles of Triangle ABC are :

• A. \(\angle A=90^\circ, \angle B=30^\circ, \angle C=60^\circ\)
• B. \(\angle A=\angle B=\angle C=60^\circ\)
• C. \(\angle A=\angle B=45^\circ, \angle C=90^\circ\)
• D. None of these

Hint:

Use Pythagoras: if \(a^2 + b^2 = c^2\), triangle is right-angled.

Answer 1

The Correct Option is D

Concept: Use distance formula to find side lengths.
Step 1: \[ \vec{AB} = (3,0,3), \quad \vec{BC} = (1,2,-7), \quad \vec{CA} = (-4,-2,4) \]
Step 2: \[ AB = \sqrt{3^2 + 0^2 + 3^2} = 3\sqrt{2} \] \[ BC = \sqrt{1^2 + 2^2 + (-7)^2} = \sqrt{54} = 3\sqrt{6} \] \[ CA = \sqrt{(-4)^2 + (-2)^2 + 4^2} = \sqrt{36} = 6 \]
Step 3: \[ AB^2 + AC^2 = 18 + 36 = 54 = BC^2 \Rightarrow \angle A = 90^\circ \]
Step 4: \[ \text{Triangle sides } \Rightarrow 3\sqrt{2}, 6, 3\sqrt{6} \] This is not a \(45^\circ\!-\!45^\circ\!-\!90^\circ\) triangle.