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Misc 16 - Find plane passing through P, perpendicular to OP

Misc 16 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2

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Misc 16 If O be the origin and the coordinates of P be (1, 2, – 3), then find the equation of the plane passing through P and perpendicular to OP.Equation of plane passing through (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is A(x − x1) + B(y − y1) + C(z − z1) = 0 The plane passes through P(1, 2, −3) So, x1 = 1, y1 = 2, z = −3 Normal vector to plane = (𝑂𝑃) ⃗ where O(0, 0, 0), P (1, 2, −3) Direction ratios of (𝑂𝑃) ⃗ = 1 − 0 , 2 − 0 , −3 − 0 = 1 , 2 , –3 ∴ A = 1, B = 2, C = −3 Equation of plane in Cartesian form is 1(x − 1) + 2 (y − 2) + (−3) (z − (−3)) = 0 x − 1 + 2y − 4 − 3 (z + 3) = 0 x − 1 + 2y − 4 − 3z − 9 = 0 x + 2y − 3z − 14 = 0

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.