<p>The direction ratios of the line <span class="math">\(x - y + z - 5 = 0 = x - 3y - 6\)</span> are</p>
<p>(a) 3, 1, −2</p>
<p>(b) 2, −4, 1</p>
<p>(c) <span class="math">\(\frac{3}{14}, \frac{1}{14}, \frac{-2}{14}\)</span></p>
<p>(d) <span class="math">\(\frac{2}{21}, \frac{-4}{21}, \frac{1}{21}\)</span></p>
Step-by-Step Solution
Key Concept: The direction of a line formed by two intersecting planes is perpendicular to both plane normals. Find the cross product of the normal vectors.
Solution: Let the DR's of the line be a, b, c. Since the line is the intersection of two planes, it is perpendicular to the normal vectors of both planes. For plane \(x - y + z - 5 = 0\) : normal vector is (1, −1, 1) For plane \(x - 3y - 6 = 0\) : normal vector is (1, −3, 0) Therefore: \(a - b + c = 0\) and \(a - 3b + 0 \cdot c = 0\) Solving these equations gives the direction ratios as (3, 1, −2), which when normalized gives \(\left(\frac{3}{14}, \frac{1}{14}, \frac{-2}{14}\right)\) . ∴ Answers are (a) and (c).
Correct Answer: A, C