Miscellaneous

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

### Transcript

Question 4 Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane π β.(π Μ + 2π Μ β 5π Μ) + 9 = 0 . The vector equation of a line passing through a point with position vector π β and parallel to vector π β is π β = π β + ππ β Given, the line passes (1, 2, 3) So, π β = 1π Μ + 2π Μ + 3π Μ Finding normal of plane π β.(π Μ + 2π Μ β 5π Μ) + 9 = 0 π β. (π Μ + 2π Μ β 5π Μ) = β 9 βπ β. (π Μ + 2π Μ β 5π Μ) = 9 π β. (β1π Μ β 2π Μ + 5π Μ) = 9 Comparing with π β. π β = d, π β = βπ Μ β 2π Μ + 5π Μ Since line is perpendicular to plane, the line will be parallel to the normal of the plane β΄ π β = π β = β1π Μ β 2π Μ + 5π Μ Hence, π β = (1π Μ + 2π Μ + 3π Μ) + π(β1π Μ β 2π Μ + 5π Μ) π β = (1π Μ + 2π Μ + 3π Μ) β π(1π Μ + 2π Μ β 5π Μ) π β = (π Μ + 2π Μ + 3π Μ) + k (π Μ + 2π Μ β 5π Μ)