Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

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Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at May 29, 2023 by Teachoo

Question 18 Find the distance between the point P(6, 5, 9) and the plane determined by the points A (3, β 1, 2), B (5, 2, 4) and C(β 1, β 1, 6).The equation of a plane passing through points A(π₯_1, π¦_1, π§_1) B(π₯_2, π¦_2, π§_2) and C (π₯_3, π¦_3, π§_3) is |β 8(πβπ_π&πβπ_π&πβπ_π@π_πβπ_π&π_πβπ_π&π_πβπ_π@π_πβπ_π&π_πβπ_π&π_πβπ_π )| = 0 Given, the three points are A(3, β1, 2) π₯_1= 3, π¦_1= β1, π§_1= 2 B(5, 2, 4) π₯_2= 5, π¦_2 = 2, π§_2= 4 C(β1, β1, 6) π₯_3= β1, π¦_3= β1, π§_3= 6 Equation of plane is |β 8(π₯β3&π¦β(β1)&π§β[email protected]β3&2β(β1)&4β2@β1β3&β1β(β1)&6β2)| = 0 |β 8(π₯β3&π¦+1&π§β[email protected]&3&2@β4&0&4)| = 0 (x β 3)[(3Γ4)β(0Γ2)] β (y + 1) [(2Γ4)β(β4Γ2)] + (z β 2) [(2Γ0)β(β4Γ3)] (x β 3)[12β0] β (y + 1) [8+8] +(π§β2) [0+12] = 0 12(x β 3) β 16(y + 1) + 12(z β 2) = 0 3(x β 3) β 4(y + 1) + 3(z β 2) = 0 3x β 9 β 4y β 4 + 3z β 6 = 0 3x β 4y + 3z β 19 = 0 Therefore, equation of plane is 3x β 4y + 3z = 19 Now, the distance between a point P(π₯_1, π¦_1, π§_1) and the plane Ax + By + Cz = D is |(π¨π_π + π©π_π + πͺπ_πβ π«)/β(π¨^π + π©^π + πͺ^π )| Given, the point is P(6, 5, 9) So, π₯_1= 6 , π¦_1= 5, π§_1= 9 The equation of plane is 3x β 4y + 3z = 19 Comparing with Ax + By + Cz = D, A = 3, B = β4, C = 3, D = 19 Now, Distance of the point from the plane = |((3 Γ 6) + (β4 Γ 5) + (3 Γ 9) β 19)/β(3^2 + 4^2 + 3^2 )| =|(18 β 20 + 27 β 19)/β(9 + 16 + 9)| =|6/β34| = 6/β34 = 6/β34 Γ β34/β34 = (6β34)/34 = (πβππ)/ππ