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Ex 11.3
Ex 11.3, 1 (b)
Ex 11.3, 1 (c) Important
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Ex 11.3, 5 (a) Important You are here
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Ex 11.3, 12 Important Deleted for CBSE Board 2022 Exams
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Last updated at Aug. 24, 2021 by Teachoo
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Ex 11.3, 5 (Introduction) Find the vector and cartesian equations of the planes (a) that passes through the point (1, 0, β2) and the normal to the plane is π Μ + π Μ β π Μ.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 Ex 11.3, 5 Find the vector and Cartesian equations of the planes (a) that passes through the point (1, 0, β 2) and the normal to the plane is π Μ + π Μ β 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, 0, β2) So, π β = 1π Μ + 0π Μ β 2π Μ Normal to plane = π Μ + π Μ β π Μ π β = π Μ + π Μ β π Μ Vector equation of plane is (π β β π β) . π β = 0 [π ββ("1" π Μ+"0" π Μβ"2" π Μ)].(π Μ + π Μ β π Μ) = 0 [π ββ(π Μβ"2" π Μ)] . (π Μ+π Μβπ Μ) = 0 Now, Finding Cartesian form using two methods Cartesian form (Method 1) Vector equation is [π ββ(π Μβ"2" π Μ)] . (π Μ+π Μβπ Μ) = 0 Put π β = xπ Μ + yπ Μ + zπ Μ [("x" π Μ+"y" π Μ+"z" π Μ )β("1" π Μ+"0" π Μβ"2" π Μ)].(1π Μ + 1π Μ β 1π Μ) = 0 [("x" β1) π Μ+("y" β0) π½ Μ+(π§β(β2))π Μ ].(1π Μ + 1π Μ β 1π Μ) = 0 1(x β 1) + 1(y β 0) β 1 (z + 2) = 0 x β 1 + y β z β 2 = 0 x + y β z = 3 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, 0, β2) x1 = 1, y1 = 0, z1 = 2 And normal is π Μ + π Μ β π Μ So, Direction ratios of line perpendicular to plane = 1, 1, β1 β΄ A = 1, B = 1, C = β1 Therefore, equation of line in Cartesian form is 1(x β 1) + 1(y β 0) β 1(z β (β2)) = 0 x + y β z = 3