Angle Between Two Planes (Cartesian Form)

In three-dimensional geometry, understanding the angle between two planes is crucial for solving various problems related to 3D space. When we talk about the angle between two planes, we refer to the angle between their normal vectors. The Cartesian form of a plane's equation is given by:

$$ Ax + By + Cz + D = 0 $$

where ( A, B, C ) are the direction ratios of the normal to the plane, and ( D ) is the constant term.

Finding the Angle Between Two Planes

To find the angle between two planes, we first need to find the normal vectors of the planes. If we have two planes with equations:

$$ Plane\ 1: A_1x + B_1y + C_1z + D_1 = 0 $$ $$ Plane\ 2: A_2x + B_2y + C_2z + D_2 = 0 $$

The normal vectors to these planes are ( \vec{n}_1 = \langle A_1, B_1, C_1 \rangle ) and ( \vec{n}_2 = \langle A_2, B_2, C_2 \rangle ), respectively.

The angle ( \theta ) between the two planes is the same as the angle between their normal vectors. This angle can be found using the dot product formula:

$$ \cos(\theta) = \frac{\vec{n}_1 \cdot \vec{n}_2}{|\vec{n}_1||\vec{n}_2|} $$

where ( \vec{n}_1 \cdot \vec{n}_2 ) is the dot product of the normal vectors, and ( |\vec{n}_1| ) and ( |\vec{n}_2| ) are the magnitudes of the normal vectors.

The dot product of the normal vectors is given by:

$$ \vec{n}_1 \cdot \vec{n}_2 = A_1A_2 + B_1B_2 + C_1C_2 $$

The magnitudes of the normal vectors are:

$$ |\vec{n}_1| = \sqrt{A_1^2 + B_1^2 + C_1^2} $$ $$ |\vec{n}_2| = \sqrt{A_2^2 + B_2^2 + C_2^2} $$

Substituting these into the cosine formula, we get:

$$ \cos(\theta) = \frac{A_1A_2 + B_1B_2 + C_1C_2}{\sqrt{A_1^2 + B_1^2 + C_1^2} \sqrt{A_2^2 + B_2^2 + C_2^2}} $$

To find the actual angle ( \theta ), we take the inverse cosine (arccos) of the result:

$$ \theta = \arccos\left(\frac{A_1A_2 + B_1B_2 + C_1C_2}{\sqrt{A_1^2 + B_1^2 + C_1^2} \sqrt{A_2^2 + B_2^2 + C_2^2}}\right) $$

Important Points and Differences

Aspect	Description
Normal Vector	A vector perpendicular to the plane. It defines the orientation of the plane in 3D space.
Dot Product	A scalar value that is the product of the magnitudes of two vectors and the cosine of the angle between them.
Angle Between Planes	The angle between the normal vectors of the planes, which can range from 0° to 180°.
Orthogonal Planes	If the angle between two planes is 90°, the planes are said to be orthogonal (perpendicular).
Parallel Planes	If the normal vectors are scalar multiples of each other, the planes are parallel.

Example

Let's find the angle between the following two planes:

$$ Plane\ 1: 2x - 3y + z - 4 = 0 $$ $$ Plane\ 2: x + y - 2z + 3 = 0 $$

First, we identify the normal vectors:

$$ \vec{n}_1 = \langle 2, -3, 1 \rangle $$ $$ \vec{n}_2 = \langle 1, 1, -2 \rangle $$

Next, we calculate the dot product and the magnitudes:

$$ \vec{n}_1 \cdot \vec{n}_2 = (2)(1) + (-3)(1) + (1)(-2) = 2 - 3 - 2 = -3 $$ $$ |\vec{n}_1| = \sqrt{2^2 + (-3)^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14} $$ $$ |\vec{n}_2| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6} $$

Now, we find the cosine of the angle:

$$ \cos(\theta) = \frac{-3}{\sqrt{14}\sqrt{6}} $$

Finally, we calculate the angle ( \theta ):

$$ \theta = \arccos\left(\frac{-3}{\sqrt{14}\sqrt{6}}\right) $$

Using a calculator, we find that ( \theta \approx 101.54° ). Therefore, the angle between Plane 1 and Plane 2 is approximately 101.54 degrees.