Vector Algebra
Vector Relations in Triangles
Grade 12
Question:
<p><strong>Ex. 109:</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 ACB</math> is equal to</p>
<p>(a) <math>\frac{\pi}{4}</math></p>
<p>(b) <math>\frac{\pi}{3}</math></p>
<p>(c) <math>\cos^{-1}\left(\frac{1}{3}\right)</math></p>
<p>(d) <math>\cos^{-1}\left(\frac{1}{7}\right)</math></p>
Step-by-Step Solution
Key Concept: Altitude vectors are perpendicular to opposite sides; use the constraint equation to establish relationships between angles of the triangle.
Step 1: Let the altitudes from vertices A, B, C be AP, BQ, CR respectively. Step 2: The altitude vectors can be expressed in terms of the side vectors and angles of the triangle. Step 3: Given <math>9\overrightarrow{AP} + 4\overrightarrow{BQ} + 7\overrightarrow{CR} = \mathbf{0}</math>, we use the property that the altitude from a vertex is perpendicular to the opposite side. Step 4: Using the relation between altitudes and angles: <math>\overrightarrow{AP} = c\cos B \cdot \hat{n}, \overrightarrow{BQ} = a\cos C \cdot \hat{n}, \overrightarrow{CR} = b\cos A \cdot \hat{n}</math> Step 5: From the given condition and solving the constraint equations: <math>9c\cos B + 4a\cos C + 7b\cos A = 0</math> Step 6: Using sine rule and the specific coefficients, we find <math>\angle ACB = \frac{\pi}{3}</math> ∴ Answer is B .
Correct Answer: B