3D Geometry
Direction cosines and angle bisectors
Grade 12
Question:
<p>The direction cosines of the lines bisecting the angle between the line whose direction cosines are <span class="math">\(l_1, m_1, n_1\)</span> and <span class="math">\(l_2, m_2, n_2\)</span> and the angle between these lines is <span class="math">\(\theta\)</span>, are</p>
<p>(a) <span class="math">\(\frac{l_1 + l_2}{\cos\frac{\theta}{2}}, \frac{m_1 + m_2}{\cos\frac{\theta}{2}}, \frac{n_1 + n_2}{\cos\frac{\theta}{2}}\)</span></p>
<p>(b) <span class="math">\(\frac{l_1 + l_2}{2\cos\frac{\theta}{2}}, \frac{m_1 + m_2}{2\cos\frac{\theta}{2}}, \frac{n_1 + n_2}{2\cos\frac{\theta}{2}}\)</span></p>
<p>(c) <span class="math">\(\frac{l_1 + l_2}{\cos\theta}, \frac{m_1 + m_2}{\cos\theta}, \frac{n_1 + n_2}{\cos\theta}\)</span></p>
Step-by-Step Solution
Key Concept: The direction of the angle bisector is the sum of the unit direction vectors. The normalization factor is the magnitude of this sum, which equals 2cos(θ/2).
Solution: The direction cosines of the angle bisectors between two lines with direction cosines \((l_1, m_1, n_1)\) and \((l_2, m_2, n_2)\) are given by the sum of these direction cosines, normalized appropriately. Since both lines are unit vectors (direction cosines), the bisector direction is proportional to \((l_1 + l_2, m_1 + m_2, n_1 + n_2)\) . The magnitude of this vector equals \(2\cos\frac{\theta}{2}\) , where \(\theta\) is the angle between the lines. Therefore, the direction cosines are \(\frac{l_1 + l_2}{2\cos\frac{\theta}{2}}, \frac{m_1 + m_2}{2\cos\frac{\theta}{2}}, \frac{n_1 + n_2}{2\cos\frac{\theta}{2}}\) . ∴ Answer is (b).
Correct Answer: B