Examples

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

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Example 6 Find the vector and the Cartesian equations of the line through the point (5, 2, β 4) and which is parallel to the vector 3π Μ + 2π Μ β 8π Μ . Vector equation Equation of a line passing through a point with position vector π β , and parallel to a vector π β is π β = π β + ππ β Since line passes through (5, 2, β4) π β = 5π Μ + 2π Μ β 4π Μ Since line is parallel to 3π Μ + 2π Μ β 8π Μ π β = 3π Μ + 2π Μ β 8π Μ Equation of line π β = π β + ππ β π β = (5π Μ + 2π Μ β 4π Μ) + π (3π Μ + 2π Μ β 8π Μ) Therefore, equation of line in vector form is π β = (5π Μ + 2π Μ β 4π Μ) + π (3π Μ + 2π Μ β 8π Μ) Cartesian equation Equation of a line passing through a point (x, y, z) and parallel to a line with direction ratios a, b, c is (π₯ β π₯1)/π = (π¦ β π¦1)/π = (π§ β π§1)/π Since line passes through (5, 2, β4) π₯1 = 5, y1 = 2 , z1 = β4 .....................................