Question

If \( A(1,0,1) \), \( B(0,1,-1) \), \( C(-1,1,0) \) are the vertices of a triangle ABC, then \( \cos^{2}A+\cos^{2}B = \)

• A. \( \frac{1}{2\sqrt{3}} \)
• B. \( \frac{1}{3\sqrt{2}} \)
• C. \( \frac{5}{6} \)
• D. \( \frac{7}{9} \)

Hint:

Recognizing that a triangle is isosceles (\( b = c \)) immediately tells you that \( \angle B = \angle C \). This symmetry reduces the number of separate side calculations needed to find the angles.

Answer 1

The Correct Option is D

Concept: The cosine of an interior angle of a triangle can be calculated using the dot product of the two vectors expanding from that vertex, or via the Law of Cosines after finding all three side lengths.

Step 1: Calculating the lengths of all three sides.
\[ a^2 = BC^2 = (-1 - 0)^2 + (1 - 1)^2 + (0 - (-1))^2 = 1 + 0 + 1 = 2 \implies a = \sqrt{2} \] \[ b^2 = AC^2 = (-1 - 1)^2 + (1 - 0)^2 + (0 - 1)^2 = 4 + 1 + 1 = 6 \implies b = \sqrt{6} \] \[ c^2 = AB^2 = (0 - 1)^2 + (1 - 0)^2 + (-1 - 1)^2 = 1 + 1 + 4 = 6 \implies c = \sqrt{6} \] Notice that \( b = c \), which means \( \triangle ABC \) is an isosceles triangle with \( \angle B = \angle C \).

Step 2: Finding \( \cos A \) and \( \cos B \) using the Law of Cosines.
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{6 + 6 - 2}{2(\sqrt{6})(\sqrt{6})} = \frac{10}{12} = \frac{5}{6} \] \[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{2 + 6 - 6}{2(\sqrt{2})(\sqrt{6})} = \frac{2}{2\sqrt{12}} = \frac{1}{2\sqrt{3}} \]

Step 3: Summing the squares of the cosines.
\[ \cos^2 A = \left(\frac{5}{6}\right)^2 = \frac{25}{36} \] \[ \cos^2 B = \left(\frac{1}{2\sqrt{3}}\right)^2 = \frac{1}{12} = \frac{3}{36} \] \[ \cos^2 A + \cos^2 B = \frac{25 + 3}{36} = \frac{28}{36} = \frac{7}{9} \] This precisely tracks Option (D).