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Chapter 11 Class 12 Three Dimensional Geometry
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Example 13 Find the vector equation of the plane which is at a distance of 6/β29 from the origin and its normal vector from the origin is 2π Μ β 3π Μ + 4π Μ.Vector equation of a plane at a distance βdβ from the origin and unit vector to normal from origin π Μ is π β.π Μ = d Unit vector of π β = π Μ = 1/|π β | (π β) Now, distance from origin = d = 6/β29 π β = 2π Μ β 3π Μ + 4π Μ Magnitude of π Μ = β(22+(β3)^2+4^2 ) |π β | = β(4+9+16) = β29 Now, π Μ = 1/|π β | " (" π β")" = 1/β29 " (2" π Μβ"3" π Μ+"4" π Μ")" = 2/β29 π Μ β 3/β29 π Μ + 4/β29 π Μ Vector equation of plane is π β.π Μ = d π β . (π/βππ π Μβπ/βππ " " π Μ+ π/βππ " " π Μ ) = π/βππ Cartesian equation Equation of a plane in Cartesian form which is at a distance βdβ from the origin and has a normal vector π β = ππ Μ + bπ Μ + cπ Μ is lx + my + nz = d where l, m, n are direction cosines of π β l = π/β(π^2 + π^2 + π^2 ) , m = π/β(π^2 + π^2 + π^2 ) , n = π/β(π^2 + π^2 + π^2 ) Distance form origin = d = 6/β29 π β = 2π Μ β 3π Μ + 4π Μ Comparing with π β = ππ Μ + ππ Μ + ππ Μ, π = 2, b = β3, c = 4 Also, β(π2+π2+π2) = β(22+(β3)2+42) = β(4+9+16) = β29 So, direction cosines are l = 2/β29 , m = ( β3)/β29 , n = 4/β29 β΄ Cartesian equations of plane is lx + my + nz = d 2/β29 x + (( β3)/β29)y + 4/β29 z = 6/β29 2x β 3y + 4z = 6 Therefore, the equation of plane in Cartesian form is 2x β 3y + 4z = 6