Equation of plane - Prependicular to Vector & Passing Through Point
Equation of plane - Prependicular to Vector & Passing Through Point
Last updated at April 16, 2024 by Teachoo
Question 7 (introduction) Find the vector and cartesian equations of the plane which passes through the point (5, 2, β 4) and perpendicular to the line with direction ratios 2, 3, β 1.Vector equation of a plane passing through a point (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is [π β β(π₯1π Μ + π¦1π Μ + π§1π Μ)]. (Aπ Μ + Bπ Μ + Cπ Μ) = 0 or (π β β π β).π β = 0 ("A" π) β is perpendicular to "n" β So, ("A" P) β . "n" β = 0 ("r" β β "a" β)."n" β = 0 Question 7 Find the vector and Cartesian equations of the plane which passes through the point (5, 2, β 4) and perpendicular to the line with direction ratios 2, 3, β 1.Vector form 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 (5, 2, β4) So π β = 5π Μ + 2π Μ β 4π Μ Direction ratios of line perpendicular to plane = 2, 3, β1 So, "n" β = 2π Μ + 3π Μ β 1π Μ Equation of plane in vector form is (π β β π β) . π β = 0 [π ββ(ππ Μ+ππ Μβππ Μ)]. (ππ Μ+ππ Μβπ Μ) = 0 Now, Finding Cartesian form using two methods Cartesian form (Method 1) Vector equation is [π ββ(5π Μ+2π Μβ4π Μ)]. (2π Μ+3π Μβπ Μ) = 0 Put π β = xπ Μ + yπ Μ + zπ Μ [(π₯π Μ+π¦π Μ+π§π Μ )β(5π Μ+2π Μβ4π Μ)].(2π Μ + 3π Μ β π Μ) = 0 [(π₯β5) π Μ+(π¦β2) π Μ+ (π§β(β 4))π Μ ].(2π Μ + 3π Μ β π Μ) = 0 2(x β 5) + 3 (y β 2) + (β 1)(z + 4) = 0 2x β 10 + 3y β 6 β z β 4 = 0 2x + 3y β z β 20 = 0 2x + 3y β z = 20 Therefore equation of plane in Cartesian form is 2x + 3y β z = 20 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 (5, 2, β4) x1 = 5, y1 = 2, z1 = β4 Direction ratios of line perpendicular to plane = 2, 3, β1 β΄ A = 2, B = 3, C = β1 Therefore, equation of line in Cartesian form is 2(x β 5) + 3 (y β 2) + (β1) (x β (β4)) = 0 2 (x β 5) + 3(y β 2) β 1 (z + 4) = 0 2x β 10 + 3y β 6 β z β 4 = 0 2x + 3y β z β 20 = 0 2x + 3y β z = 20 Therefore equation of plane in Cartesian form is 2x + 3y β z = 20