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Question 15 - Angle between Line and Plane - Chapter 11 Class 12 Three Dimensional Geometry
Last updated at Dec. 16, 2024 by Teachoo
Angle between Line and Plane
Chapter 11 Class 12 Three Dimensional Geometry
Concept wise
- Direction cosines and ratios
- Equation of line - given point and //vector
- Equation of line - given 2 points
- Angle between two lines - Vector
- Angle between two lines - Cartisian
- Angle between two lines - Direction ratios or cosines
- Shortest distance between two skew lines
- Shortest distance between two parallel lines
- Equation of plane - In Normal Form
- Equation of plane - Prependicular to Vector & Passing Through Point
- Equation of plane - Passing Through 3 Non Collinear Points
- Equation of plane - Intercept Form
- Equation of plane - Passing Through Intersection Of Planes
- Coplanarity of 2 lines
- Angle between two planes
- Distance of point from plane
-
Angle between Line and Plane
- Equation of line under planes condition
- Point with Lines and Planes
Transcript
Question 15 Find the angle between the line (π₯ + 1)/2 = π¦/3 = (π§ β 3)/6 And the plane 10x + 2y β 11z = 3. The angle between a line (π₯ β π₯_1)/π = (π¦ β π¦_1)/π = (π§ βγ π§γ_1)/π and the normal to the plane Ax + By + Cz = D is given by cos ΞΈ = |(π΄π + π΅π + πΆπ)/(β(π^2 + π^2 +γ πγ^2 ) β(π΄^2 +γ π΅γ^2 +γ πΆγ^2 ))| So, angle between line and plane is given by sin π = |(π΄π + π΅π + πΆπ)/(β(π^2 + π^2 + π^2 )+β(π΄^2 + π΅^2 +γ πΆγ^2 ))| Given, the line is (π₯ + 1)/2 = π¦/3 = (π§ β 3)/6 (π₯ β (β1))/2 = (π¦ β 0)/3 = (π§ β 3)/6 Comparing with (π₯ βγ π₯γ_1)/π = (π¦ β π¦_1)/π = (π§ β π§_1)/π , π = 2, b = 3, c = 6 The plane is 10x + 2y β 11z = 3 Comparing with Ax + By + Cz = D, A = 10, B = 2, C = β11 So, sin Ο = |((10 Γ 2) + (2 Γ 3) + (β11 Γ 6))/(β(2^2 + 3^2 + 6^2 ) β(γ10γ^(2 )+γ 2γ^2 + γ(β11)γ^2 ))| = |(20 + 6 β 66)/(β(4 + 9 + 36) β(100 + 4 + 121))| = |(β40)/(7 Γ 15)| = 8/21 So, sin Ο = 8/21 β΄ π = γπππγ^(βπ)β‘(π/ππ) Therefore, the angle between the given line and plane is sin^(β1)β‘(8/21).
