Question

If \( A(1, 2, -3) \), \( B(2, 3, -1) \), and \( C(3, 1, 1) \) are the vertices of triangle \( \Delta ABC \), then find \( \left| \frac{\cos A}{\cos B} \right|.\)

• A. \( \frac{3\sqrt{3}}{4\sqrt{2}} \)
• B. \( \frac{3\sqrt{3}}{\sqrt{7}} \)
• C. \( \frac{4\sqrt{2}}{3\sqrt{3}} \)
• D. \( \frac{\sqrt{7}}{3\sqrt{3}} \) \bigskip

Hint:

When finding angles in a triangle using vectors, always first find the vector components, calculate the dot product, and then compute the magnitudes before applying the formula for cosine of the angle.

Answer 1

The Correct Option is B

We are given the points \( A(1, 2, -3) \), \( B(2, 3, -1) \), and \( C(3, 1, 1) \), and we need to find \( \left| \frac{\cos A}{\cos B} \right| \). Step 1: Find the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) The vector \( \overrightarrow{AB} \) is given by: \[ \overrightarrow{AB} = B - A = (2 - 1, 3 - 2, -1 - (-3)) = (1, 1, 2) \] The vector \( \overrightarrow{AC} \) is given by: \[ \overrightarrow{AC} = C - A = (3 - 1, 1 - 2, 1 - (-3)) = (2, -1, 4) \] Step 2: Use the dot product to find \( \cos A \) The formula for \( \cos A \) is: \[ \cos A = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}| |\overrightarrow{AC}|} \] First, compute the dot product \( \overrightarrow{AB} \cdot \overrightarrow{AC} \): \[ \overrightarrow{AB} \cdot \overrightarrow{AC} = (1)(2) + (1)(-1) + (2)(4) = 2 - 1 + 8 = 9 \] Now, compute the magnitudes of \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \): \[ |\overrightarrow{AB}| = \sqrt{(1)^2 + (1)^2 + (2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \] \[ |\overrightarrow{AC}| = \sqrt{(2)^2 + (-1)^2 + (4)^2} = \sqrt{4 + 1 + 16} = \sqrt{21} \] Thus, \[ \cos A = \frac{9}{\sqrt{6} \times \sqrt{21}} = \frac{9}{\sqrt{126}} = \frac{9}{\sqrt{9 \times 14}} = \frac{3}{\sqrt{14}} \] Step 3: Use the dot product to find \( \cos B \) Next, compute the vector \( \overrightarrow{BC} \): \[ \overrightarrow{BC} = C - B = (3 - 2, 1 - 3, 1 - (-1)) = (1, -2, 2) \] Now, use the formula for \( \cos B \): \[ \cos B = \frac{\overrightarrow{BC} \cdot \overrightarrow{AC}}{|\overrightarrow{BC}| |\overrightarrow{AC}|} \] Compute the dot product \( \overrightarrow{BC} \cdot \overrightarrow{AC} \): \[ \overrightarrow{BC} \cdot \overrightarrow{AC} = (1)(2) + (-2)(-1) + (2)(4) = 2 + 2 + 8 = 12 \] Now, compute the magnitude of \( \overrightarrow{BC} \): \[ |\overrightarrow{BC}| = \sqrt{(1)^2 + (-2)^2 + (2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \] Thus, \[ \cos B = \frac{12}{3 \times \sqrt{21}} = \frac{12}{3\sqrt{21}} = \frac{4}{\sqrt{21}} \] Step 4: Find \( \left| \frac{\cos A}{\cos B} \right| \) Now, we can compute \( \left| \frac{\cos A}{\cos B} \right| \): \[ \left| \frac{\cos A}{\cos B} \right| = \left| \frac{\frac{3}{\sqrt{14}}}{\frac{4}{\sqrt{21}}} \right| \] \[ \left| \frac{\cos A}{\cos B} \right| = \left| \frac{3}{\sqrt{14}} \times \frac{\sqrt{21}}{4} \right| \] \[ \left| \frac{\cos A}{\cos B} \right| = \frac{3}{4} \times \frac{\sqrt{21}}{\sqrt{14}} = \frac{3}{4} \times \sqrt{\frac{21}{14}} = \frac{3}{4} \times \sqrt{\frac{3}{2}} = \frac{3\sqrt{3}}{4\sqrt{2}} \] Thus, the final answer is \( \frac{\sqrt{3}}{\sqrt{7}} \). \bigskip