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# Angle Between Two Planes

Coordinate geometry is a special branch of mathematics, in which we learn about the coordinates or the locations of the geometrical objects in Euclidean space. The concepts that are being utilized very frequently in coordinate geometry are - points, lines, angles, planes, spaces, circles, parabola, hyperbola, ellipse, graphs etc.

Let us discuss about planes here. A plane is another name of a surface which has no width. In other words, a plane is defined as a two-dimensional piece of flat surface. It is said to be an
analogue of a line (which is in one dimension) into two dimensions. The planes are also used as the subspaces of a higher-dimensional space.

In mathematics, we may come across with problems having two or more planes. The planes may intersect one another. In three dimensions, two different planes must be either parallel or Intersect in the form of a line segment.

When two planes intersect, an angle is formed between them as shown in the following diagram:

This angle may be denoted by $\alpha$ or even by $\theta$. The angle between two planes shown in above figure is also known as the dihedral angle.

In this article, we are going to go ahead and understand about the angle between two planes, its formula, its calculations and solved examples based on this concept.

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## Formula

Eventually, there are two angles that can be formed between two planes - one is acute and another is obtuse; while in the case of perpendicular planes, both angles are equal. The angle between two planes determines the measure of the acute angle. It is done by the use of normal vectors of the given planes.

Let us first recall what the normal vector means. The vector which is normal or perpendicular to a plane, is called its normal vector.

Have a look at following diagram for better understanding:

In this diagram, $\prod$ denotes the given plane and $\vec{n}$ is its normal vector.
Numerically, If a plane is represented by the following equation:

ax + by + cz + d = 0, then its normal vector will be (a, b, c).

Let us suppose that the equations of two intersecting planes are given by:

$\prod_{1} : a_{1} x + b_{1} y + c_{1} z + d_{1} = 0$

and

$\prod_{2} : a_{2} x + b_{2} y + c_{2} z + d_{2} = 0$

Then, the angle between these planes (i.e. the angle between their normal vectors) is calculated by the following formula:
$cos \alpha$ = $\frac{|\vec{n_{1}} . \vec{n_{2}}|}{\vec{n_{1}}\ \vec{n_{2}}}$
Where, $\alpha$ = acute angle between two planes.

$\vec{n_{1}}$ = normal vector to plane $\prod_{1}$ = ($a_{1}, b_{1}, c_{1}$)

$\vec{n_{2}}$ = normal vector to plane $\prod_{2}$ = ($a_{2}, b_{2}, c_{2}$)

Thus, above formula can attain the following form:
$cos \alpha$ = $\frac{|a_{1} a_{2} + b_{1} b_{2} + c_{1}c_{2}|}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$

Obtuse angle = $180^{\circ} - \alpha$.

## How to Find?

In order to find the angle between two planes, the following mentioned steps are to be followed:
Step 1: Observe the equations of planes if they are in the form $a_{1} x + b_{1} y + c_{1} z + d_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} z + d_{2} = 0$. If not rearrange and get this form.

Step 2: Find the normal vectors corresponding to each plane which are ($a_{1}, b_{1}, c_{1}$) and ($a_{2}, b_{2}, c_{2}$) respectively.

Step 3: Substitute these values in the formula:
$cos \alpha$ = $\frac{|a_{1} a_{2} + b_{1} b_{2} + c_{1}c_{2}|}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$

Step 4: Solve and get the value of $\alpha$
$\alpha$ = $cos^{-1} [\frac{|a_{1} a_{2} + b_{1} b_{2} + c_{1}c_{2}|}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}]$

Step 5: If another angle is also asked in the question, then subtract the measure of $\alpha$ by $180^{\circ}$ to obtain this angle.

## Examples

The examples based on calculation of angle between two planes are as follows:

Example 1: Find the angle between the planes x + y - z - 7 = 0 and x + y + z + 9 = 0.

Solution: Normal vector of plane
x + y - z - 7 = 0 are (1, 2, -3)

Normal vector of plane 2 x + y + z + 9 = 0 are (2, 4, 6)

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

and $a_{2}$ =
2, $b_{2}$ = 4, $c_{2}$ = 6

$cos \alpha$ = $\frac{|a_{1} a_{2} + b_{1} b_{2} + c_{1}c_{2}|}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$

$cos \alpha$ = $\frac{|1 \times 2 + 2 \times 4 + (-3) \times 6|}{\sqrt{1^{2} + 2^{2} + (-3)^{2}} \sqrt{2^{2}+4^{2}+6^{2}}}$

$cos \alpha$ = $\frac{|2 + 8 - 18|}{\sqrt{1 + 4 + 9} \sqrt{4+16+36}}$

$cos \alpha$ = $\frac{8}{\sqrt{14} \sqrt{56}}$

$cos \alpha$ = $\frac{8}{\sqrt{784}}$

$cos \alpha$ = $\frac{8}{28}$

$cos \alpha$ = $\frac{2}{7}$

$\alpha$ = $cos^{-1} \frac{2}{7}$

Example 2: Calculate the measure of angle between the planes 2x -  y + z - 1 = 0 and x + z + 3 = 0.

Solution: Normal vector for plane 2x -  y + z - 1 = 0 are (2, -1, 1)

Normal vector for plane x + z + 3 = 0 are (1, 0, 1)

Thus, $a_{1}$ = 2, $b_{1}$ = -1, $c_{1}$ = 1

and $a_{2}$ = 1, $b_{2}$ = 0, $c_{2}$ = 1

$cos \alpha$ = $\frac{|a_{1} a_{2} + b_{1} b_{2} + c_{1}c_{2}|}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$

$cos \alpha$ = $\frac{|2 \times 1 + (-1) \times 0 + 1 \times 1|}{\sqrt{2^{2} + (-1)^{2} + 1^{2}} \sqrt{1^{2}+0^{2}+1^{2}}}$

$cos \alpha$ = $\frac{|2 + 0 + 1|}{\sqrt{4 + 1 + 1} \sqrt{1+ 0 + 1}}$

$cos \alpha$ = $\frac{3}{\sqrt{6} \sqrt{2}}$

$cos \alpha$ = $\frac{3}{\sqrt{12}}$

$cos \alpha$ = $\frac{3}{2 \sqrt{3}}$

$cos \alpha$ = $\frac{\sqrt{3}}{2}$

$\alpha$ = $30^{\circ}$
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