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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Mathematics » Angle between Two Planes

Angle between Two Planes

If we have two planes and we are required to measure the angle in between, then the concept of the angle between two planes comes into existence.

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In general, dealing with 2-dimensional figures and problems is very common and can usually be solved by drawing. But understanding 3-dimensional figures is a little bit uncommon because we can’t draw every diagram properly on paper. So, we need to understand it mentally ( hypothetically). We can assume one of the corners of our room as the origin and the attached edges as X, Y, and Z axes, accordingly. This can help easily frame any diagram in our minds. Now, we can easily see that there is an angle of 90⁰ between each edge (what we had considered at the room’s corner). This was easy. But when it comes to measuring a random angle between two planes, it becomes quite difficult. In fact, we can’t directly measure it (as we can do in 2-dimensional figures). So, to measure the angle between two planes, we need to first understand the terms “direction cosine” and “direction ratio”.

Direction Cosine 

Let us come to the corner of our room. Suppose we draw any line vector OZ on the floor. Now, the angle measured by this line from the X, Y, and Z axes in an anti-clockwise direction is the direction angle along the vector OZ. Let us suppose these angles are α, β, and γ, respectively. Then, the cosine of these angles, i.e., cosα, cosβ, and cosγ are the direction cosine and represented as l, m, and n, respectively ( in standard form).

● Direction cosines are always unique.

● l²+m²+n²=1

Direction Ratio

Suppose we have 3 variables: a, b and c. Any a, b, and c satisfying 

l/a = m/b = n/c are always direction ratios.

● Direction ratios are not unique

Now, basically, the angle between two planes is equal to the angle between their normal. And now, all our struggle turns in the direction to find the angle between normal planes, which is quite easy. We can find it either by vector plane method or by cartesian plane method (basically, both come from one another, but the condition is what is given in the question and what they have asked for).

1- Vector plane method 

Till now, we have come to know that the angle between two planes is just the same as the angle between the normal of those planes. 

Now suppose we have two equations in vector form as-

r . n1 – d₁ =0……………..(i)

&, r. n1 – d₂=0………………(ii)

Where ⃗n₁ and ⃗n₂ are the normal vectors and d₁  and d₂  are the distance of the plane from the origin, respectively.

Then, the angle between two planes n₁ and n₂ is 

Cosθ = n₁.n₂        

          |n₁| |n₂|

2-Cartesian plane method 

From equation (i) ,                 

r . n1 =d₁

And equation in the cartesian plane is-

a₁x+ b₁y + c₁z +d₁=0

On comparing, 

n₁= a₁ i + b₁ j + c₁ k 

      

This implies the direction ratio will be < a₁ b₁ c₁>.

Similarly, the direction ratio for the 2nd plane will be <a₂ b₂ c₂>.

Now,

Cosθ =<a₁ b₁ c₁> <a₂ b₂ c₂>

√(a₁²+b₁²+c₁²(a₂²+b₂²+c₂²)

          = a₁a₂  + b₁b₂  + c₁c₂  √(a₁²+b₁²+c₁²)√(a₂²+b₂²+c)

Now, if we have equations given in the cartesian plane, then we don’t need to go for the direction ratio.

As If we have given equations as,

a₁x+ b₁y +c₁z +d₁=0

& a₂x + b₂y + c₂z +d₂=0

Where a₁, b₁, c₁ are the direction ratio of normal to the 1st plane and a₂, b₂, c₂ are direction ratio of normal to the 2nd plane.

Then, the angle between the two planes is

       Cosθ= a₁a₂+ b₁b₂+ c₁c₂ 

√(a₁²+b₁²+c₁²(a₂²+b₂²+c₂²)

■ Case1- if planes are perpendicular-

This means angle between planes is 90⁰

This gives,

Cos90⁰ = a₁a₂ + b₁b₂ + c₁c₂ 

√ (a₁²+b₁²+c₁²) . √ (a₂²+b₂²+c₂²)

It implies    0 = a₁a₂ + b₁b₂ + c₁c₂  ( as cos 90⁰=0)

√ (a₁²+b₁²+c₁²) . √ (a₂²+b₂²+c₂²)

So, a₁a₂ + b₁b₂+ c₁c₂ =0

■Case2- if planes are parallel-

This means the angle between planes is 0⁰.

This gives, 

Cos0⁰ =  a₁a₂  + b₁b₂  + c₁c₂

√ (a₁²+b₁²+c₁²) . √ (a₂²+b₂²+c₂²)

It implies, 

a₁a₂ +b₁b₂+ c₁c₂ =√ (a₁²+b₁²+c₁²)√ (a₂²+b₂²+c₂²)

(Since cos0⁰=1)

This gives,           

a₁/a₂ = b₁/b₂ = c₁/c₂

Conclusion-

We can find the angle between two planes by measuring the angle between their normal. And the general formula for measuring the angle between two planes is       

Cosθ = n₁  .n₂

          |n₁| |n₂|

Where n₁ and n₂  are normal vectors of planes 1 and 2, respectively.

Generally, questions regarding angles between 2 planes are based on calculations. It requires lots of practice. The angle between 2 planes notes is beneficial for every entrance exam and it is an important topic too.

faq

Frequently Asked Questions

Get answers to the most common queries related to the JEE Examination Preparation.

Find angle between two planes 2x + 3y+ 5z +6=0 and 3x +4y+z-1=0 .

Ans : As we know, if we have two equations a₁x+b₁y+c₁z+d₁=0 and a₂x+b₂y+c₂z+d₂=0. Then, the angle be...Read full

Find the angle between two planes x+5y=0 and y+2z=3?

Ans : As we know, if we have two equations a₁x+b₁y+c₁z+d₁=0 and a₂x+b₂y+c₂z+d₂=0. Then angle between...Read full

Find the angle between two planes x+ 2y+ 3z=0 and 3x - 4z=0 using the vector method?

Ans :We have a standard form of 3D equations as r ...Read full

Suppose we have two equations-3x + 4y - z +5=0 and 2x+y -4=0. Find the smallest angle between two planes?

Ans : As we know, if we have two equations, a₁x+b₁y+c₁z+d₁=0 and a₂x+b₂y+c₂z+d₂=0. Then angle betwee...Read full

Ans : As we know, if we have two equations a₁x+b₁y+c₁z+d₁=0 and a₂x+b₂y+c₂z+d₂=0. Then, the angle between two planes is, 

        Cosθ = a₁a₂ + b₁b₂ +c₁c₂

                    √ (a₁²+b₁²+c₁²).√ (a₂²+b₂²+c₂²)

On comparing,  a₁=2, b₁=3, c₁=5

                         a₂=3,b₂=4,c₂=1

              Cosθ=  2.3 + 3.4 + 5.1 

                         √ (2²+3²+5²).√ (3²+4²+1²)

                     = 23

 2.√ (13×19)

              θ = cos inverse  23

                                      2.√ (13×19)

Ans : As we know, if we have two equations a₁x+b₁y+c₁z+d₁=0 and a₂x+b₂y+c₂z+d₂=0. Then angle between two planes is, 

                          Cosθ = a₁a₂ + b₁b₂ +c₁c₂

                    √ (a₁²+b₁²+c₁²).√ (a₂²+b₂²+c₂²)

On comparing,  a₁=1, b₁=5, c₁=0

                         a₂=0, b₂=1, c₂=2

   It gives, cos θ= 1.0 + 5.1 + 0.2 

                           √ (1²+5²+0²).√ (0²+1²+2²)

                      =5/ √ (26×5)

                θ= cos inverse √ (5/26)

Ans :We have a standard form of 3D equations as r . n1 =d₁

                                                               And,       r . n2 = d₂

Given equations in vector form is r ( i+2j+3k) =0

                                                          r (3i-4k)=0

 We know angle between two planes in vector form is  

                     Cosθ=   n₁ .n₂

                                |n₁| |n₂|

                           = <1 2 3 > <3 0 -4>

                              √ (1²+2²+3²).√ (3²+0²+(-4)²)

                           =  3+0-12

                              √ (14×25)

                        = (-9/5√ (14))

    It implies,  θ = cos inverse (-9/ 5√ (14))

Ans : As we know, if we have two equations, a₁x+b₁y+c₁z+d₁=0 and a₂x+b₂y+c₂z+d₂=0. Then angle between two planes is,     

        Cosθ = a₁a₂ + b₁b₂ +c₁c₂

                    √ (a₁²+b₁²+c₁²).√ (a₂²+b₂²+c₂²)

  On comparing, a₁= -3, b₁= 4 , c₁ = -1

                         a₂= 2, b₂= 1, c₂=0

         Cosθ =  (-3).2 + 4.1 + (-1).0 

                     √ ((-3)²+4²+(-1)²)√ (2²+1²+0²)

               = (-2)/ √ (5×26)

            θ = cos inverse (-2)/ √ (5×26)

For the smallest angle, neglect negative sign,

It implies, 

             θ = cos inverse 2/ √ (5×26)

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