If angle BCD is \theta, then what is \cos^{2}\theta equal to?
For the following two (02) items: Let A(1,-1,0), B(-2,1,8) and C(-1,2,7) are three consecutive vertices of a parallelogram ABCD.
- A. 26/77
- B. 27/77 ✓
- C. 82/237
- D. 83/237
Correct Answer: B. 27/77
Explanation
Angle \theta is between vectors \vec{CB} and \vec{CD}. We have \vec{CB} = B - C = \langle -1, -1, 1 \rangle and \vec{CD} = D - C = \langle 3, -2, -8 \rangle. The dot product is (-1)(3) + (-1)(-2) + (1)(-8) = -9. Their magnitudes are |\vec{CB}| = \sqrt{3} and |\vec{CD}| = \sqrt{9+4+64} = \sqrt{77}. Thus, \cos\theta = \frac{-9}{\sqrt{3}\sqrt{77}}, meaning \cos^2\theta = \frac{81}{231} = \frac{27}{77}.
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