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 (180^{0}- θ) = angle between the two planes (Fig (b)).

r^{->}.n_{1}⃗=d_{1} and r^{->}.n_{2}⃗=d Where n_{1}⃗ and n_{2}⃗ = normal to the planes, θ =Angle between the planes.

θ= angle between the normal to the planes drawn from some common point.

Therefore,

** Note: -**The planes are perpendicular to each other if n

__Cartesian Form:-__

Let θ =angle between the planes.

A_{1}x+B_{1}y+C_{1}z+D_{1}=0 and A_{2}x+B_{2}y+C_{2}z+D_{2}=0 where A_{1},B_{1},C_{1} and A_{2},B_{2},C_{2}

Therefore ,

__Note: - __

If the planes are at right angles, then θ=90^{0} and so cosθ =0. Therefore ,

cosθ= (A_{1}A_{2 }+ B_{1}B_{2}+ C_{1}C_{2}).

If the planes are parallel, then (A_{1}/A_{2}) = (B_{1}/B_{2}) = (C_{1}/C_{2}).

__Problem:-__

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.

** Answer**:-

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 n_{1}⃗ and n_{2}⃗ are r^{->}.n_{1}⃗=d_{1} and r⃗.n_{2}⃗=d_{2}normal to the planes, then the angle between them, θ, is given by,

Here, n_{1}⃗= (2î+2ĵ-3k̂) and n_{2}⃗= (3î - 3ĵ+5k̂)

Therefore, n_{1}⃗.n_{2}⃗= (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

A_{1} x + B_{1} y + C_{1} z + D_{1} = 0 and A_{2} x + B_{2} y + C_{2} z + D_{2} = 0

We get A_{1} = 3, B_{1} = – 6, C_{1} = 2 A_{2} = 2, B_{2} = 2, C_{2} = – 2

= (5)/ (7√3)

= (5√3)/ (21)

Therefore, θ = cos^{-1} ((5√3)/ (21))

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