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

To find the angles of triangle \(ABC\), with vertices \(A(-1, 3, 2)\), \(B(2, 3, 5)\), and \(C(3, 5, -2)\), we will employ the use of the dot product of vectors. The dot product can help us determine the cosine of the angles at the vertices of the triangle.

Step 1: Determine the Vectors

First, we determine the vectors corresponding to the sides of the triangle:

• \(\overrightarrow{AB} = B - A = (2 - (-1), 3 - 3, 5 - 2) = (3, 0, 3)\)
• \(\overrightarrow{BC} = C - B = (3 - 2, 5 - 3, -2 - 5) = (1, 2, -7)\)
• \(\overrightarrow{CA} = A - C = (-1 - 3, 3 - 5, 2 - (-2)) = (-4, -2, 4)\)

Step 2: Calculate Magnitudes of Vectors

Next, we calculate the magnitudes of these vectors:

• \(|\overrightarrow{AB}| = \sqrt{3^2 + 0^2 + 3^2} = \sqrt{18} = 3\sqrt{2}\)
• \(|\overrightarrow{BC}| = \sqrt{1^2 + 2^2 + (-7)^2} = \sqrt{54} = 3\sqrt{6}\)
• \(|\overrightarrow{CA}| = \sqrt{(-4)^2 + (-2)^2 + 4^2} = \sqrt{36} = 6\)

Step 3: Calculate the Dot Products

Now, let's calculate the dot products, which will allow us to use the cosine rule:

• \(\overrightarrow{AB} \cdot \overrightarrow{BC} = 3 \times 1 + 0 \times 2 + 3 \times (-7) = 3 - 21 = -18\)
• \(\overrightarrow{BC} \cdot \overrightarrow{CA} = 1 \times (-4) + 2 \times (-2) + (-7) \times 4 = -4 - 4 - 28 = -36\)
• \(\overrightarrow{CA} \cdot \overrightarrow{AB} = (-4) \times 3 + (-2) \times 0 + 4 \times 3 = -12 + 12 = 0\)

Step 4: Calculate Angles using Cosine Formula

We can use the cosine formula \(\cos \theta = \frac{\overrightarrow{u} \cdot \overrightarrow{v}}{|\overrightarrow{u}| |\overrightarrow{v}|}\) to find each angle.

For angle at \(A\) between vectors \(\overrightarrow{AB}\) and \(\overrightarrow{CA}\):

• \(\cos A = \frac{0}{|\overrightarrow{AB}| |\overrightarrow{CA}|} = 0 \Rightarrow A = 90^\circ\)

For angle at \(B\) between vectors \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\):

• \(\cos B = \frac{-18}{|\overrightarrow{AB}| |\overrightarrow{BC}|} = \frac{-18}{3 \sqrt{2} \times 3 \sqrt{6}} = \cos B = -\frac{1}{\sqrt{3}} \Rightarrow B = 150^\circ\)

For angle at \(C\) between vectors \(\overrightarrow{BC}\) and \(\overrightarrow{CA}\):

• \(\cos C = \frac{-36}{|\overrightarrow{BC}| |\overrightarrow{CA}|} = \frac{-36}{3\sqrt{6} \times 6} = -\frac{1}{\sqrt{3}} \Rightarrow C = 120^\circ\)

Conclusion

Thus, the angles of the triangle are \(A = 90^\circ\), \(B = 150^\circ\), and \(C = 120^\circ\). None of the given options match these calculated angles, confirming that the correct answer is:

None of these