



Get live Maths 1-on-1 Classs - Class 6 to 12
Examples
Example, 2 Important
Example, 3
Example, 4 Important
Example, 5 Important
Example, 6 Important
Example, 7
Example 8
Example, 9
Example 10 Important
Example 11
Example 12 Important
Example 13 Important Deleted for CBSE Board 2023 Exams You are here
Example 14 Deleted for CBSE Board 2023 Exams
Example 15 Deleted for CBSE Board 2023 Exams
Example 16 Important Deleted for CBSE Board 2023 Exams
Example 17 Deleted for CBSE Board 2023 Exams
Example 18 Deleted for CBSE Board 2023 Exams
Example 19 Important Deleted for CBSE Board 2023 Exams
Example 20 Important Deleted for CBSE Board 2023 Exams
Example 21 Important Deleted for CBSE Board 2023 Exams
Example 22 Deleted for CBSE Board 2023 Exams
Example 23 Important Deleted for CBSE Board 2023 Exams
Example 24 Deleted for CBSE Board 2023 Exams
Example, 25 Important Deleted for CBSE Board 2023 Exams
Example 26
Example 27 Important Deleted for CBSE Board 2023 Exams
Example 28 Important Deleted for CBSE Board 2023 Exams
Example 29 Important
Example 30 Important Deleted for CBSE Board 2023 Exams
Last updated at March 22, 2023 by Teachoo
Example 13 Find the vector equation of the plane which is at a distance of 6/β29 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 βdβ 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/β29 π β = 2π Μ β 3π Μ + 4π Μ Magnitude of π Μ = β(22+(β3)^2+4^2 ) |π β | = β(4+9+16) = β29 Now, π Μ = 1/|π β | " (" π β")" = 1/β29 " (2" π Μβ"3" π Μ+"4" π Μ")" = 2/β29 π Μ β 3/β29 π Μ + 4/β29 π Μ Vector equation of plane is π β.π Μ = d π β . (π/βππ π Μβπ/βππ " " π Μ+ π/βππ " " π Μ ) = π/βππ Cartesian equation Equation of a plane in Cartesian form which is at a distance βdβ from the origin and has a normal vector π β = ππ 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/β29 π β = 2π Μ β 3π Μ + 4π Μ Comparing with π β = ππ Μ + ππ Μ + ππ Μ, π = 2, b = β3, c = 4 Also, β(π2+π2+π2) = β(22+(β3)2+42) = β(4+9+16) = β29 So, direction cosines are l = 2/β29 , m = ( β3)/β29 , n = 4/β29 β΄ Cartesian equations of plane is lx + my + nz = d 2/β29 x + (( β3)/β29)y + 4/β29 z = 6/β29 2x β 3y + 4z = 6 Therefore, the equation of plane in Cartesian form is 2x β 3y + 4z = 6