- 2071 Bhadra · 5 marks
Symmetric form of line of two planes
Short answer
Direction ratios from cross product of normals (1,2,3) and (3,4,5) are (1,-2,1). Setting z=0 gives point (-10,8,0). Symmetric form: (x+10)/1 = (y-8)/(-2) = z/1.
Express the equation of the line x+2y+3z−6 = 0, 3x+4y+5z−2 = 0 in symmetric form.
Given. The line is the intersection of x + 2y + 3z − 6 = 0 and 3x + 4y + 5z − 2 = 0. To write it in symmetric form we need a point on it and its direction ratios.
Direction ratios
The line is perpendicular to both normals (1, 2, 3) and (3, 4, 5), so its direction is their cross product:
d = (1, 2, 3) × (3, 4, 5)
= (2·5 − 3·4, 3·3 − 1·5, 1·4 − 2·3)
= (−2, 4, −2) ∝ (1, −2, 1)
A point on the line
Put z = 0: x + 2y = 6 and 3x + 4y = 2. Subtract twice the first from the second: x = 2 − 12 = −10. Then y = 8. So (−10, 8, 0) lies on the line.
Check in the second plane: 3(−10) + 4(8) + 0 − 2 = 0 ✓.
Symmetric form
(x + 10)/1 = (y − 8)/(−2) = z/1
Answer: (x + 10)/1 = (y − 8)/(−2) = z/1.
✓ Answer checked by Claude Opus 5.5 (high effort)
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