Ex 11.3

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

Β

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Ex 11.3, 5 Find the vector and cartesian equations of the planes (b) That passes through the point (1, 4, 6) and the normal vector to the plane is π Μ β 2π Μ + π Μ Vector equation Equation of plane passing through point A whose position vector is π β & perpendicular to π β is (π β β π β) . π β = 0 Given, Plane passes through (1, 4, 6) So, π β = 1π Μ + 4π Μ + 6π Μ Normal to plane = π Μ β 2π Μ + π Μ π β = π Μ β 2π Μ + π Μ Vector equation of plane is (π β β π β).π β = 0 [π ββ(π Μ+ππ Μ+ππ Μ)] . (π Μ β 2π Μ + π Μ) = 0 Now, Finding Cartesian form using two methods Cartesian form (Method 1) : Vector equation is [π ββ(π Μ+4π Μ+6π Μ)] . (π Μ β 2π Μ + π Μ) = 0 Put π β = xπ Μ + yπ Μ + zπ Μ [(π₯π Μ+π¦π Μ+π§π Μ )β(1π Μ+4π Μ+6π Μ)]. (1π Μ β 2π Μ + 1π Μ) = 0 [(π₯β1) π Μ +(π¦β4) π Μ+(π§β6)π Μ ]. (1π Μ β 2π Μ + 1π Μ) = 0 1(x β 1) + (β2)(y β 4) + 1 (z β 6) = 0 x β 1 β 2(y β 4) + z β 6 = 0 x β 2y + z + 1 = 0 β΄ Equation of plane in Cartesian form is x β 2y + z + 1 = 0. Cartesian form (Method 2) Equation of plane passing through (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is A(x β x1) + B(y β y1) + C (z β z1) = 0 Since the plane passes through (1, 4, 6) x1 = 1, y1 = 4, z1 = 6 And normal is π Μ β 2π Μ + π Μ So, Direction ratios of line perpendicular to plane = 1, β2, 1 β΄ A = 1, B = β2, C = 1 Therefore, equation of line in Cartesian form is 1(x β 1) β 2 (y β 4) + 1(x β 6) = 0 x β 2y + z + 1 = 0