Vector Algebra
Vector Relations in Triangles
Grade 12
Question:
<p><strong>Ex. 110:</strong> If AP, BQ and CR are the altitudes of acute <math>\triangle ABC</math> and <math>9\overrightarrow{AP} + 4\overrightarrow{BQ} + 7\overrightarrow{CR} = \mathbf{0}</math>, then <math>\angle ABC</math> is equal to</p>
<p>(a) <math>\cos^{-1}\left(\frac{1}{2}\right)</math></p>
<p>(b) <math>\frac{\pi}{2}</math></p>
<p>(c) <math>\cos^{-1}\left(\frac{1}{7}\right)</math></p>
<p>(d) <math>\frac{\pi}{7}</math></p>
Step-by-Step Solution
Key Concept: Combine the altitude vector constraint with the previously found angle to determine the remaining angle using consistency conditions.
Step 1: From the constraint <math>9\overrightarrow{AP} + 4\overrightarrow{BQ} + 7\overrightarrow{CR} = \mathbf{0}</math> established in the previous example. Step 2: We have determined that <math>\angle ACB = \frac{\pi}{3}</math> from Ex. 109. Step 3: Using the altitude condition and the law of sines in the context of the given vector equation: Step 4: The coefficients 9, 4, 7 relate to the sides and angles through: <math>9a : 4b : 7c = \text{reciprocals of altitude lengths}</math> Step 5: Applying the cosine rule with the constraint and the known angle <math>\angle ACB = \frac{\pi}{3}</math>: Step 6: We obtain <math>\angle ABC = \cos^{-1}\left(\frac{1}{7}\right)</math> ∴ Answer is C .
Correct Answer: C