Examples

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

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

Transcript

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(πβπ_π&πβπ_π&πβπ_π@π_πβπ_π&π_πβπ_π&π_πβπ_π@π_πβπ_π&π_πβπ_π&π_πβπ_π )| = 0 Given, the three points are A(3, β1, 2) π₯_1= 3, π¦_1= β1, π§_1= 2 B(5, 2, 4) π₯_2= 5, π¦_2 = 2, π§_2= 4 C(β1, β1, 6) π₯_3= β1, π¦_3= β1, π§_3= 6 Equation of plane is |β 8(π₯β3&π¦β(β1)&π§β[email protected]β3&2β(β1)&4β2@β1β3&β1β(β1)&6β2)| = 0 |β 8(π₯β3&π¦+1&π§β[email protected]&3&2@β4&0&4)| = 0 (x β 3)[(3Γ4)β(0Γ2)] β (y + 1) [(2Γ4)β(β4Γ2)] + (z β 2) [(2Γ0)β(β4Γ3)] (x β 3)[12β0] β (y + 1) [8+8] +(π§β2) [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 + πͺπ_πβ π«)/β(π¨^π + π©^π + πͺ^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 = β4, C = 3, D = 19 Now, Distance of the point from the plane = |((3 Γ 6) + (β4 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/β34| = 6/β34 = 6/β34 Γ β34/β34 = (6β34)/34 = (πβππ)/ππ