Misc 16 - Chapter 11 Class 12 Three Dimensional Geometry

Transcript

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