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

Step 1: Find the three side lengths.
Using the 3D distance formula with $A(1,0,1)$, $B(0,1,-1)$, $C(-1,1,0)$:
Step 2: Side $a=BC$.
\[ a^2=(-1-0)^2+(1-1)^2+(0+1)^2=1+0+1=2 \]
Step 3: Sides $b=AC$ and $c=AB$.
\[ b^2=(-1-1)^2+(1-0)^2+(0-1)^2=4+1+1=6 \] \[ c^2=(0-1)^2+(1-0)^2+(-1-1)^2=1+1+4=6 \] So $b=c$, meaning the triangle is isosceles.
Step 4: Find $\cos A$ by the law of cosines.
\[ \cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{6+6-2}{2\cdot6}=\frac{10}{12}=\frac{5}{6} \]
Step 5: Find $\cos B$.
\[ \cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{2+6-6}{2\sqrt{2}\sqrt{6}}=\frac{2}{2\sqrt{12}}=\frac{1}{2\sqrt{3}} \]
Step 6: Add the squares.
\[ \cos^2A+\cos^2B=\frac{25}{36}+\frac{1}{12}=\frac{25}{36}+\frac{3}{36}=\frac{28}{36}=\frac{7}{9} \] \[ \boxed{\tfrac{7}{9}} \]