Question

Let \( A = (1, -2, 3) \), \( B = (3, 1, -3) \), \( C = (-3, 1, 3) \) be the vertices of triangle \( ABC \). If \( \cos \theta = \frac{\vec{AB} \cdot \vec{AC}}{|\vec{AB}||\vec{AC}|} \), then \( \cos A \) is:

• A. \( -\frac{1}{35} \)
• B. \( \frac{1}{7} \)
• C. \( -\frac{1}{7} \)
• D. \( \frac{1}{35} \)

Hint:

Cosine of Angle between Vectors}
Use dot product formula: \( \cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|} \)
Carefully compute vector components
Always reduce fractions at the end

Answer 1

The Correct Option is D

Vectors: \[ \vec{AB} = (2, 3, -6),\quad \vec{AC} = (-4, 3, 0) \] Dot product: \[ \vec{AB} \cdot \vec{AC} = (2)(-4) + (3)(3) + (-6)(0) = -8 + 9 = 1 \] Magnitudes: \[ |\vec{AB}| = \sqrt{4 + 9 + 36} = \sqrt{49} = 7,\quad |\vec{AC}| = \sqrt{16 + 9 + 0} = \sqrt{25} = 5 \] \[ \cos A = \frac{1}{7 \cdot 5} = \frac{1}{35} \]