|Class 12 Maths Three Dimensional Geometry||Angle between Two Planes: Vector|
Angle between Two Planes: Vector
The angle between 2 planes is defined as the angle which they make with their normal.
Let θ = angle between the two planes (Fig (a)) then (1800- θ) = angle between the two planes (Fig (b)).
r->.n1⃗=d1 and r->.n2⃗=d Where n1⃗ and n2⃗ = normal to the planes, θ =Angle between the planes.
θ= angle between the normal to the planes drawn from some common point.
Note: -The planes are perpendicular to each other if n1⃗.n2⃗=0 and parallel ifn1⃗ is parallel to n2⃗.
Let θ =angle between the planes.
A1x+B1y+C1z+D1=0 and A2x+B2y+C2z+D2=0 where A1,B1,C1 and A2,B2,C2
If the planes are at right angles, then θ=900 and so cosθ =0. Therefore ,
cosθ= (A1A2 + B1B2+ C1C2).
If the planes are parallel, then (A1/A2) = (B1/B2) = (C1/C2).
Find the angle between the planes whose vector equations are: - r->. (2 î+2ĵ-3 k̂)=5 and
r->. (3 î - 3ĵ+5 k̂)=3.
The equations of the given planes are r->. (2 î+2ĵ-3 k̂)=5 and r->. (3 î - 3ĵ+5 k̂)=3.
It is known that if n1⃗ and n2⃗ are r->.n1⃗=d1 and r⃗.n2⃗=d2normal to the planes, then the angle between them, θ, is given by,
Here, n1⃗= (2î+2ĵ-3k̂) and n2⃗= (3î - 3ĵ+5k̂)
Therefore, n1⃗.n2⃗= (2î+2ĵ-3k̂). (3î - 3ĵ+5k̂) =2.3+2. (-3)+ (-3).5=-15
Substituting the value
cosθ = (15)/ (√ (731)
Therefore θ =cos-1(15)/ (√ (731)
Problem:Find the angle between the two planes 3x – 6y + 2z = 7 and 2x + 2y – 2z =5.
Answer: - Comparing the given equations of the planes with the equations
A1 x + B1 y + C1 z + D1 = 0 and A2 x + B2 y + C2 z + D2 = 0
We get A1 = 3, B1 = – 6, C1 = 2 A2 = 2, B2 = 2, C2 = – 2
= (5)/ (7√3)
= (5√3)/ (21)
Therefore, θ = cos-1 ((5√3)/ (21))