What is the angle between two planes?

The angle between two planes is the acute angle created by the intersection of the planes. It is the angle between their normal vectors.

Is the angle between two planes always acute?

No, the angle between two planes is not always acute. It is typically taken as the smaller angle formed by their intersection, which can be acute or obtuse, but by convention, we often refer to the acute angle.

Can the angle between two planes be zero or 180 degrees?

Yes, if two planes are parallel, the angle between them is either 0 degrees (if they are identical or parallel in the same direction) or 180 degrees (if they are parallel in opposite directions).

What is the measure of a dihedral angle?

A dihedral angle is the angle between two intersecting planes. It is measured along a line perpendicular to their line of intersection.

Frequently Asked Questions

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Angle Between Two Lines

When two straight lines intersect, they create two sets of angles: one pair of acute angles and one pair of obtuse angles. The exact values of these angles depend on the slopes of the intersecting lines.

Angle Between Two Vectors

The angle formed between two vectors is referred to as the angle between the vectors formed at their tails.

Altitude and Median Triangle

Altitude and median of a triangle are key geometric concepts. The altitude is a perpendicular segment from a vertex...

Altitude of a Triangle

The Altitude of a Triangle is a perpendicular segment or line drawn from a vertex of the triangle to the line that extends through the opposite side (known as the base).

Solution of Triangles

The solution of a triangle involves finding its various elements such as angles, side lengths, and area, typically using trigonometric principles such as the law of sines and the law of cosines.

Sequence and Series

A sequence refers to a structured arrangement of elements in a specific order that follows a specific pattern or rule.

Maxima and Minima

Maxima are the highest points on a graph where the function reaches its peak and Minima is...

Introduction to 3D geometry

Three-dimensional geometry encompasses the mathematical study of shapes and figures in 3D space, involving three coordinates:

Angle Between Two Planes

The Angle Between Two Planes is defined as the acute angle created where the planes intersect. It is the angle between the normal vectors of the planes. Mathematically, if and are the normal vectors of the two planes, the angle θ between the planes can be calculated using the dot product:

$cos θ = \frac{n _{1} \cdot n _{2}}{∣ n _{1} ∣ ∣ n _{2} ∣}$

where $n_{1} \cdot n_{2}$ is the dot product of the normal vectors, and $n_{1}$ and $n_{2}$ are the magnitudes of the normal vectors. This angle is typically measured in degrees or radians.

1.0Angle Between Two Planes in Vector Form

Consider two planes intersecting each other in a straight line.

If $n_{1}$ is normal to one plane and $n_{2}$ is normal to the other plane then the planes are described by the equation $r \cdot n_{1} = d_{1}$ and $r \cdot n_{2} = d_{2}$ .

Let us assume the acute angle formed between two planes is θ . Obtuse angle formed Between two planes is 180 – θ.

So, the angle between the normal is θ. Let us now calculate angle between the 2 normal which can be calculated by the fact that the dot product of the vectors $n_{1}$ and $n_{2}$ is equal to the product of magnitude and cosine of angle θ.

$n_{1} \cdot n_{2} = n_{1} n_{2} cos θ$

$\frac{n _{1} \cdot n _{2}}{∣ n _{1} ∣ ∣ n _{2} ∣} = cos θ$

$θ = cos^{- 1} \frac{n _{1} \cdot n _{2}}{∣ n _{1} ∣ ∣ n _{2} ∣}$

Note: $n_{1} \times n_{2}$ is parallel to the intersection of 2 planes.

Important Points:

The line of intersection of planes is always perpendicular to both the normal.

If the normal $n_{1}$ and $n_{2}$ are perpendicular to each other that means they are at 90° with each other, then dot product is equal to 0.

$n_{1} \cdot n_{2} = n_{1} c d o t n_{2} cos 9 0^{\circ}$

$n_{1} \cdot n_{2} = 0 {∵ cos 9 0^{\circ} = 0}$

If the dot product of $n_{1}$ and $n_{2}$ are zero, then the planes are perpendicular to each other, and the angle defined by the planes can be defined as the angle between the normal.
If the angle between the normal $n_{1}$ and $n_{2}$ is 0° then the normal are parallel to each other.
Two planes are parallel to each other if their normal are parallel.

Summary:

• Angle between the two planes is defined as the angle between their normal.

• If θ is the angle between the two planes represented by the equations $r \cdot n_{1} = d_{1}$ and $r \cdot n_{2} = d_{2}$ , then $θ = cos^{- 1} \frac{n _{1} \cdot n _{2}}{∣ n _{1} ∣ ∣ n _{2} ∣} .$

• Two planes are perpendicular to each other if dot product of their normal is equal to zero.

• Two planes are parallel if their normal are parallel to each other.

2.0Angle Between Two Planes in Coordinate Form

Angle between the planes $a_{1} x + b_{1} y + c_{1} z + d_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} z + d_{2} = 0$ is

$cos θ = \frac{a _{1} a _{2} + b _{1} b _{2} + c _{1} c _{2}}{a _{1}^{2} + b _{1}^{2} + c _{1}^{2} a _{2}^{2} + b _{2}^{2} + c _{2}^{2}}$

If direction cosine is given, then

$cos θ = \frac{ℓ _{1} ℓ _{2} + m _{1} m _{2} + n _{1} n _{2}}{ℓ _{1}^{2} + m _{1}^{2} + n _{1}^{2} ℓ _{2}^{2} + m _{2}^{2} + n _{2}^{2}}$

$cos θ = ℓ_{1} ℓ_{2} + m_{1} m_{2} + n_{1} n_{1}$

$[ℓ_{1}^{2} + m_{1}^{2} + n_{1}^{2} = 1 ℓ_{2}^{2} + m_{2}^{2} + m_{2}^{2} = 1]$

If θ = 90°, then cos 90° = 0 (two planes are perpendicular)

∴ a₁a₂ + b₁b₂ + c₁c₂ = 0 and $ℓ_{1} ℓ_{2} + m_{1} m_{2} + n_{1} n_{2} = 0$

If θ = 0°, $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ (two planes are parallel)

3.0Solved Examples on Angle Between Two Planes

Example 1: The angle between the plane 2x – y + z = 6 and x + y + 2z = 3 is

Solution:

a₁ = 2, b₁ = –1, c₁ = 1

a₂ = 1, b₂ = 1, c₂ = 2

Comparing both equation by $a_{1} x + b_{1} y + c_{1} z + d_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} z + d_{2} = 0$

Using formula for cartesian forms

$cos θ = \frac{a _{1} a _{2} + b _{1} b _{2} + c _{1} c _{2}}{a _{1}^{2} + b _{1}^{2} + c _{1}^{2} a _{2}^{2} + b _{2}^{2} + c _{2}^{2}}$

$\Rightarrow cos θ = \frac{2 \cdot 1 + ( - 1 ) \cdot 1 + 1 \cdot 2}{2 ^{2} + ( - 1 ) ^{2} + 1 ^{2} 1 ^{2} + 1 ^{2} + 2 ^{2}}$

$\Rightarrow cos θ = \frac{2 - 1 + 2}{4 + 1 + 1 1 + 1 + 4}$

$\Rightarrow cos θ = \frac{3}{6 6}$

$\Rightarrow cos θ = \frac{3}{6}$

$\Rightarrow cos θ = \frac{1}{2}$

$\Rightarrow θ = cos^{- 1} (\frac{1}{2})$

$\Rightarrow θ = \frac{π}{3}$

Example 2: Find the angle between the planes 2x – 3y + 4z = 5 and 3x + 5y – 2z = 7

Solution:

Comparing both equations by $a_{1} x + b_{1} y + c_{1} z - d_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} z - d_{2} = 0$ .

So, $a_{1}$ =2, $b_{1}$ =-3, $c_{1}$ =4

and a2 = 3, b2 = 5, c2 = –2

Using formula for cartesian form

$cos θ = \frac{a _{1} a _{2} + b _{1} b _{2} + c _{1} c _{2}}{a _{1}^{2} + b _{1}^{2} + c _{1}^{2} a _{2}^{2} + b _{2}^{2} + c _{2}^{2}}$

$\Rightarrow cos θ = \frac{2 \cdot 3 + ( - 3 ) \cdot 5 + 4 \cdot ( - 2 )}{2 ^{2} + ( - 3 ) ^{2} + 4 ^{2} 3 ^{2} + 5 ^{2} + ( - 2 ) ^{2}}$

$\Rightarrow cos θ = \frac{6 - 15 - 8}{4 + 9 + 16 9 + 25 + 4}$

$\Rightarrow cos θ = \frac{- 17}{29 38}$

$\Rightarrow cos θ = \frac{- 17}{1102}$

$\Rightarrow θ = cos^{- 1} (\frac{- 17}{1102})$

Example 3: Find the angle between the planes 3x + y – 2z = 7 and 2x – 4y + z = 3.

Solution:

Comparing both equations by $a_{1} x + b_{1} y + c_{1} z - d_{1} = 0$ and

$a_{2} x + b_{2} y + c_{2} z - d_{2} = 0$ .

So, a₁ = 3, b₁ = 1, c₁ = –2

and a₂ = 2, b₂ = –4, c₂ = 1

Using formula for cartesian form

$cos θ = \frac{a _{1} a _{2} + b _{1} b _{2} + c _{1} c _{2}}{a _{1}^{2} + b _{1}^{2} + c _{1}^{2} a _{2}^{2} + b _{2}^{2} + c _{2}^{2}}$

$\Rightarrow cos θ = \frac{3 \cdot 2 + 1 ( - 4 ) + ( - 2 ) 1}{3 ^{2} + 1 ^{2} + ( - 2 ) ^{2} 2 ^{2} + ( - 4 ) ^{2} + 1 ^{2}}$

$\Rightarrow cos θ = \frac{6 - 4 - 2}{9 + 1 + 4 4 + 16 + 1}$

$\Rightarrow cos θ = \frac{0}{14 21}$

$\Rightarrow cos θ = 0$

$\Rightarrow θ = cos^{- 1} (0)$

$\Rightarrow θ = 9 0^{\circ}$

So, the angle between the planes is $9 0^{\circ}$ .

Example 4: Find the angle between the planes 2x + 3y – z = 5 and x – 4y + 2z = –6.

Solution:

For the plane 2x + 3y – z = 5, the normal vector $n_{1} = ⟨ 2, 3, - 1 ⟩$

For the plane x – 4y + 2z = –6, the normal vector $n_{2} = ⟨ 1, - 4, 2 ⟩$

Using formula

$cos θ = \frac{n _{1} \cdot n _{2}}{∣ n _{1} ∣ ∣ n _{2} ∣}$

$\Rightarrow cos θ = \frac{2 \cdot 1 + 3 ( - 4 ) + ( - 1 ) \cdot 2}{2 ^{2} + 3 ^{2} + ( - 1 ) ^{2} 1 ^{2} + ( - 4 ) ^{2} + 2 ^{2}}$

$\Rightarrow cos θ = \frac{2 - 12 - 2}{4 + 9 + 1 1 + 16 + 4}$

$\Rightarrow cos θ = \frac{- 12}{14 21}$

$\Rightarrow cos θ = \frac{- 12}{29}$

$\Rightarrow cos θ = \frac{- 12}{7 6}$

$\Rightarrow θ = cos^{- 1} (\frac{- 12}{7 6})$

Example 5: Find the angle between the planes x – 2y + 3z = 4 and 4x + y – z = 2.

Solution:

For the plane x – 2y + 3z = 4, the normal vector $n_{1} = ⟨ 1, - 2, 3 ⟩$

For the plane 4x + y – z = 2, the normal vector $n_{2} = ⟨ 4, 1, - 1 ⟩$

Dot Product of both normal are:

$n_{1} \cdot n_{2}$ = (1)(4) + (–2)(1) + (3)(–1)

= 4 – 2 – 3 = 1

Magnitude of normal are:

$∣ n_{1} ∣ = 1^{2} + (- 2)^{2} + 3^{2} = 1 + 4 + 9 = 14$

$n_{2} = 4^{2} + 1^{2} + (- 1)^{2} = 16 + 1 + 1 = 18 = 32$

Using formula

$cos θ = \frac{n _{1} \cdot n _{2}}{∣ n _{1} ∣ \cdot ∣ n _{2} ∣}$

$\Rightarrow cos θ = \frac{- 1}{14 \cdot 3 2}$

$\Rightarrow cos θ = \frac{- 1}{3 28}$

$\Rightarrow cos θ = \frac{- 1}{3 \cdot 2 7}$

$\Rightarrow cos θ = \frac{- 1}{6 7}$

$\Rightarrow θ = cos^{- 1} (\frac{- 1}{6 7})$

4.0Practice Questions on Angle Between Two Planes

1. Find the angle between the planes 3x + 2y – z = 0 and x – y + z = 1.

2. Find the angle between the planes 2x + 3y – z = 5 and x – 4y + 2z = –6.

3. Find the angle between the planes 4x + y + z = 1 and x – y + z = 2.

4. Find the angle between the planes 3x + 2y – z = 6 and x – 4y + 2z = 7.

5. Find the angle between the planes 5x – 3y + 2z = 8 and 3x + y – z = 4.

5.0Frequently Asked Questions on Angle Between Two Planes

1. How do you find the angle between two planes?

Ans: To find the angle between two planes, you can use the dot product of their normal vectors. If $n_{1}$ and $n_{2}$ are the normal vectors of the planes, the angle q can be calculated using: $cos θ = \frac{n _{1} \cdot n _{2}}{∣ n _{1} ∣ ∣ n _{2} ∣}$ . Then, $θ = cos^{- 1} (cos θ) .$

2. What are normal vectors of a plane?

Ans: Normal vectors are vectors that are perpendicular to the plane. For a plane given by ax + by + cz = d, the normal vector is $n = ⟨ a, b, c ⟩$ .

3. How do you find the direction cosines of a normal vector?

Ans: Direction cosines are the cosines of the angles that a vector makes with the coordinate axes. For a vector $n = ⟨ a, b, c ⟩$ , the direction cosines are:

$l = \frac{a}{∣ n ∣}, m = \frac{b}{∣ n ∣}, n = \frac{c}{∣ n ∣}$ , where, |n|= $a^{2} + b^{2} + c^{2}$ .

Join ALLEN!

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Angle Between Two Lines

When two straight lines intersect, they create two sets of angles: one pair of acute angles and one pair of obtuse angles. The exact values of these angles depend on the slopes of the intersecting lines.

Angle Between Two Vectors

The angle formed between two vectors is referred to as the angle between the vectors formed at their tails.

Altitude and Median Triangle

Altitude and median of a triangle are key geometric concepts. The altitude is a perpendicular segment from a vertex...

Altitude of a Triangle

The Altitude of a Triangle is a perpendicular segment or line drawn from a vertex of the triangle to the line that extends through the opposite side (known as the base).

Solution of Triangles

The solution of a triangle involves finding its various elements such as angles, side lengths, and area, typically using trigonometric principles such as the law of sines and the law of cosines.

Sequence and Series

A sequence refers to a structured arrangement of elements in a specific order that follows a specific pattern or rule.

Maxima and Minima

Maxima are the highest points on a graph where the function reaches its peak and Minima is...

Introduction to 3D geometry

Three-dimensional geometry encompasses the mathematical study of shapes and figures in 3D space, involving three coordinates:

Angle Between Two Planes

The Angle Between Two Planes is defined as the acute angle created where the planes intersect. It is the angle between the normal vectors of the planes. Mathematically, if and are the normal vectors of the two planes, the angle θ between the planes can be calculated using the dot product:

$cos θ = \frac{n _{1} \cdot n _{2}}{∣ n _{1} ∣ ∣ n _{2} ∣}$

where $n_{1} \cdot n_{2}$ is the dot product of the normal vectors, and $n_{1}$ and $n_{2}$ are the magnitudes of the normal vectors. This angle is typically measured in degrees or radians.

1.0Angle Between Two Planes in Vector Form

Consider two planes intersecting each other in a straight line.

If $n_{1}$ is normal to one plane and $n_{2}$ is normal to the other plane then the planes are described by the equation $r \cdot n_{1} = d_{1}$ and $r \cdot n_{2} = d_{2}$ .

Let us assume the acute angle formed between two planes is θ . Obtuse angle formed Between two planes is 180 – θ.

So, the angle between the normal is θ. Let us now calculate angle between the 2 normal which can be calculated by the fact that the dot product of the vectors $n_{1}$ and $n_{2}$ is equal to the product of magnitude and cosine of angle θ.

$n_{1} \cdot n_{2} = n_{1} n_{2} cos θ$

$\frac{n _{1} \cdot n _{2}}{∣ n _{1} ∣ ∣ n _{2} ∣} = cos θ$

$θ = cos^{- 1} \frac{n _{1} \cdot n _{2}}{∣ n _{1} ∣ ∣ n _{2} ∣}$

Note: $n_{1} \times n_{2}$ is parallel to the intersection of 2 planes.

Important Points:

The line of intersection of planes is always perpendicular to both the normal.

If the normal $n_{1}$ and $n_{2}$ are perpendicular to each other that means they are at 90° with each other, then dot product is equal to 0.

$n_{1} \cdot n_{2} = n_{1} c d o t n_{2} cos 9 0^{\circ}$

$n_{1} \cdot n_{2} = 0 {∵ cos 9 0^{\circ} = 0}$

If the dot product of $n_{1}$ and $n_{2}$ are zero, then the planes are perpendicular to each other, and the angle defined by the planes can be defined as the angle between the normal.
If the angle between the normal $n_{1}$ and $n_{2}$ is 0° then the normal are parallel to each other.
Two planes are parallel to each other if their normal are parallel.

Summary:

• Angle between the two planes is defined as the angle between their normal.

• If θ is the angle between the two planes represented by the equations $r \cdot n_{1} = d_{1}$ and $r \cdot n_{2} = d_{2}$ , then $θ = cos^{- 1} \frac{n _{1} \cdot n _{2}}{∣ n _{1} ∣ ∣ n _{2} ∣} .$

• Two planes are perpendicular to each other if dot product of their normal is equal to zero.

• Two planes are parallel if their normal are parallel to each other.

2.0Angle Between Two Planes in Coordinate Form

Angle between the planes $a_{1} x + b_{1} y + c_{1} z + d_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} z + d_{2} = 0$ is

$cos θ = \frac{a _{1} a _{2} + b _{1} b _{2} + c _{1} c _{2}}{a _{1}^{2} + b _{1}^{2} + c _{1}^{2} a _{2}^{2} + b _{2}^{2} + c _{2}^{2}}$

If direction cosine is given, then

$cos θ = \frac{ℓ _{1} ℓ _{2} + m _{1} m _{2} + n _{1} n _{2}}{ℓ _{1}^{2} + m _{1}^{2} + n _{1}^{2} ℓ _{2}^{2} + m _{2}^{2} + n _{2}^{2}}$

$cos θ = ℓ_{1} ℓ_{2} + m_{1} m_{2} + n_{1} n_{1}$

$[ℓ_{1}^{2} + m_{1}^{2} + n_{1}^{2} = 1 ℓ_{2}^{2} + m_{2}^{2} + m_{2}^{2} = 1]$

If θ = 90°, then cos 90° = 0 (two planes are perpendicular)

∴ a₁a₂ + b₁b₂ + c₁c₂ = 0 and $ℓ_{1} ℓ_{2} + m_{1} m_{2} + n_{1} n_{2} = 0$

If θ = 0°, $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ (two planes are parallel)

3.0Solved Examples on Angle Between Two Planes

Example 1: The angle between the plane 2x – y + z = 6 and x + y + 2z = 3 is

Solution:

a₁ = 2, b₁ = –1, c₁ = 1

a₂ = 1, b₂ = 1, c₂ = 2

Comparing both equation by $a_{1} x + b_{1} y + c_{1} z + d_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} z + d_{2} = 0$

Using formula for cartesian forms

$cos θ = \frac{a _{1} a _{2} + b _{1} b _{2} + c _{1} c _{2}}{a _{1}^{2} + b _{1}^{2} + c _{1}^{2} a _{2}^{2} + b _{2}^{2} + c _{2}^{2}}$

$\Rightarrow cos θ = \frac{2 \cdot 1 + ( - 1 ) \cdot 1 + 1 \cdot 2}{2 ^{2} + ( - 1 ) ^{2} + 1 ^{2} 1 ^{2} + 1 ^{2} + 2 ^{2}}$

$\Rightarrow cos θ = \frac{2 - 1 + 2}{4 + 1 + 1 1 + 1 + 4}$

$\Rightarrow cos θ = \frac{3}{6 6}$

$\Rightarrow cos θ = \frac{3}{6}$

$\Rightarrow cos θ = \frac{1}{2}$

$\Rightarrow θ = cos^{- 1} (\frac{1}{2})$

$\Rightarrow θ = \frac{π}{3}$

Example 2: Find the angle between the planes 2x – 3y + 4z = 5 and 3x + 5y – 2z = 7

Solution:

Comparing both equations by $a_{1} x + b_{1} y + c_{1} z - d_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} z - d_{2} = 0$ .

So, $a_{1}$ =2, $b_{1}$ =-3, $c_{1}$ =4

and a2 = 3, b2 = 5, c2 = –2

Using formula for cartesian form

$cos θ = \frac{a _{1} a _{2} + b _{1} b _{2} + c _{1} c _{2}}{a _{1}^{2} + b _{1}^{2} + c _{1}^{2} a _{2}^{2} + b _{2}^{2} + c _{2}^{2}}$

$\Rightarrow cos θ = \frac{2 \cdot 3 + ( - 3 ) \cdot 5 + 4 \cdot ( - 2 )}{2 ^{2} + ( - 3 ) ^{2} + 4 ^{2} 3 ^{2} + 5 ^{2} + ( - 2 ) ^{2}}$

$\Rightarrow cos θ = \frac{6 - 15 - 8}{4 + 9 + 16 9 + 25 + 4}$

$\Rightarrow cos θ = \frac{- 17}{29 38}$

$\Rightarrow cos θ = \frac{- 17}{1102}$

$\Rightarrow θ = cos^{- 1} (\frac{- 17}{1102})$

Example 3: Find the angle between the planes 3x + y – 2z = 7 and 2x – 4y + z = 3.

Solution:

Comparing both equations by $a_{1} x + b_{1} y + c_{1} z - d_{1} = 0$ and

$a_{2} x + b_{2} y + c_{2} z - d_{2} = 0$ .

So, a₁ = 3, b₁ = 1, c₁ = –2

and a₂ = 2, b₂ = –4, c₂ = 1

Using formula for cartesian form

$cos θ = \frac{a _{1} a _{2} + b _{1} b _{2} + c _{1} c _{2}}{a _{1}^{2} + b _{1}^{2} + c _{1}^{2} a _{2}^{2} + b _{2}^{2} + c _{2}^{2}}$

$\Rightarrow cos θ = \frac{3 \cdot 2 + 1 ( - 4 ) + ( - 2 ) 1}{3 ^{2} + 1 ^{2} + ( - 2 ) ^{2} 2 ^{2} + ( - 4 ) ^{2} + 1 ^{2}}$

$\Rightarrow cos θ = \frac{6 - 4 - 2}{9 + 1 + 4 4 + 16 + 1}$

$\Rightarrow cos θ = \frac{0}{14 21}$

$\Rightarrow cos θ = 0$

$\Rightarrow θ = cos^{- 1} (0)$

$\Rightarrow θ = 9 0^{\circ}$

So, the angle between the planes is $9 0^{\circ}$ .

Example 4: Find the angle between the planes 2x + 3y – z = 5 and x – 4y + 2z = –6.

Solution:

For the plane 2x + 3y – z = 5, the normal vector $n_{1} = ⟨ 2, 3, - 1 ⟩$

For the plane x – 4y + 2z = –6, the normal vector $n_{2} = ⟨ 1, - 4, 2 ⟩$

Using formula

$cos θ = \frac{n _{1} \cdot n _{2}}{∣ n _{1} ∣ ∣ n _{2} ∣}$

$\Rightarrow cos θ = \frac{2 \cdot 1 + 3 ( - 4 ) + ( - 1 ) \cdot 2}{2 ^{2} + 3 ^{2} + ( - 1 ) ^{2} 1 ^{2} + ( - 4 ) ^{2} + 2 ^{2}}$

$\Rightarrow cos θ = \frac{2 - 12 - 2}{4 + 9 + 1 1 + 16 + 4}$

$\Rightarrow cos θ = \frac{- 12}{14 21}$

$\Rightarrow cos θ = \frac{- 12}{29}$

$\Rightarrow cos θ = \frac{- 12}{7 6}$

$\Rightarrow θ = cos^{- 1} (\frac{- 12}{7 6})$

Example 5: Find the angle between the planes x – 2y + 3z = 4 and 4x + y – z = 2.

Solution:

For the plane x – 2y + 3z = 4, the normal vector $n_{1} = ⟨ 1, - 2, 3 ⟩$

For the plane 4x + y – z = 2, the normal vector $n_{2} = ⟨ 4, 1, - 1 ⟩$

Dot Product of both normal are:

$n_{1} \cdot n_{2}$ = (1)(4) + (–2)(1) + (3)(–1)

= 4 – 2 – 3 = 1

Magnitude of normal are:

$∣ n_{1} ∣ = 1^{2} + (- 2)^{2} + 3^{2} = 1 + 4 + 9 = 14$

$n_{2} = 4^{2} + 1^{2} + (- 1)^{2} = 16 + 1 + 1 = 18 = 32$

Using formula

$cos θ = \frac{n _{1} \cdot n _{2}}{∣ n _{1} ∣ \cdot ∣ n _{2} ∣}$

$\Rightarrow cos θ = \frac{- 1}{14 \cdot 3 2}$

$\Rightarrow cos θ = \frac{- 1}{3 28}$

$\Rightarrow cos θ = \frac{- 1}{3 \cdot 2 7}$

$\Rightarrow cos θ = \frac{- 1}{6 7}$

$\Rightarrow θ = cos^{- 1} (\frac{- 1}{6 7})$

4.0Practice Questions on Angle Between Two Planes

1. Find the angle between the planes 3x + 2y – z = 0 and x – y + z = 1.

2. Find the angle between the planes 2x + 3y – z = 5 and x – 4y + 2z = –6.

3. Find the angle between the planes 4x + y + z = 1 and x – y + z = 2.

4. Find the angle between the planes 3x + 2y – z = 6 and x – 4y + 2z = 7.

5. Find the angle between the planes 5x – 3y + 2z = 8 and 3x + y – z = 4.

5.0Frequently Asked Questions on Angle Between Two Planes

1. How do you find the angle between two planes?

Ans: To find the angle between two planes, you can use the dot product of their normal vectors. If $n_{1}$ and $n_{2}$ are the normal vectors of the planes, the angle q can be calculated using: $cos θ = \frac{n _{1} \cdot n _{2}}{∣ n _{1} ∣ ∣ n _{2} ∣}$ . Then, $θ = cos^{- 1} (cos θ) .$

2. What are normal vectors of a plane?

Ans: Normal vectors are vectors that are perpendicular to the plane. For a plane given by ax + by + cz = d, the normal vector is $n = ⟨ a, b, c ⟩$ .

3. How do you find the direction cosines of a normal vector?

Ans: Direction cosines are the cosines of the angles that a vector makes with the coordinate axes. For a vector $n = ⟨ a, b, c ⟩$ , the direction cosines are:

$l = \frac{a}{∣ n ∣}, m = \frac{b}{∣ n ∣}, n = \frac{c}{∣ n ∣}$ , where, |n|= $a^{2} + b^{2} + c^{2}$ .

Frequently Asked Questions

What is the angle between two planes?

Is the angle between two planes always acute?

Can the angle between two planes be zero or 180 degrees?

What is the measure of a dihedral angle?

Frequently Asked Questions

What is the angle between two planes?

Is the angle between two planes always acute?

Can the angle between two planes be zero or 180 degrees?

What is the measure of a dihedral angle?

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Angle Between Two Planes

1.0Angle Between Two Planes in Vector Form

2.0Angle Between Two Planes in Coordinate Form

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Table of Contents

Join ALLEN!

Related Articles:-

Angle Between Two Lines

Angle Between Two Vectors

Altitude and Median Triangle

Altitude of a Triangle

Solution of Triangles

Sequence and Series

Maxima and Minima

Introduction to 3D geometry

Angle Between Two Planes

1.0Angle Between Two Planes in Vector Form

2.0Angle Between Two Planes in Coordinate Form

3.0Solved Examples on Angle Between Two Planes

4.0Practice Questions on Angle Between Two Planes

5.0Frequently Asked Questions on Angle Between Two Planes

Table of Contents