Parallel, perpendicular and angle between planes

Partial Derivatives - Calculus 3

 
Parallel, perpendicular and angle between planes
 

Given two planes:

 
Parallel, perpendicular and angle between planes
 

they will always be

parallel if the ratio equality is true.

???\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}???

perpendicular if the dot product of their normal vectors is ???0???.

???a\cdot{b}=0???

set at a non-???90^\circ??? angle if the planes are neither parallel nor perpendicular, in which case the angle between the planes is given by

???\cos{\theta}=\frac{a\cdot{b}}{|a||b|}???

where ???a??? and ???b??? are the normal vectors to the given planes, ???a\cdot{b}??? is the dot product of the vectors, ???|a|??? is the magnitude of the vector ???a??? (its length) and ???|b|??? is the magnitude of the vector ???b??? (its length). We can find the magnitude of both vectors using the distance formula

???D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}???

for a three-dimensional vector where the point ???(x_1,y_1,z_1)??? is the origin ???(0,0,0)???.

 
 

Let’s look at an example.


Example

Say whether the planes are parallel, perpendicular, or neither. If the planes are neither parallel nor perpendicular, find the angle between the planes.

???3x-y+2z=5???

???x+4y+3z=1???

 

First we’ll find the normal vectors of the given planes.

 
Parallel, perpendicular and angle between planes
 

To say whether the planes are parallel, we’ll set up our ratio inequality using the direction numbers from their normal vectors.

???\frac31=\frac{-1}{4}=\frac23???

Since the ratios are not equal, the planes are not parallel.

To say whether the planes are perpendicular, we’ll take the dot product of their normal vectors.

???a\cdot{b}=(3)(1)+(-1)(4)+(2)(3)???

???a\cdot{b}=3-4+6???

???a\cdot{b}=5???

Since the dot product is not ???0???, the planes are not perpendicular.

Since the planes are not parallel or perpendicular, we know that they are set at a non-???90^\circ??? angle from one another, which is given by the formula

???\cos{\theta}=\frac{a\cdot{b}}{|a||b|}???

We need to find the dot product of the normal vectors, and the magnitude of each of them. We already know from our perpendicular test that their dot product is

???a\cdot{b}=(3)(1)+(-1)(4)+(2)(3)???

???a\cdot{b}=3-4+6???

???a\cdot{b}=5???

The magnitude of ???a\langle 3,-1,2\rangle??? is

???|a|=\sqrt{(3-0)^2+(-1-0)^2+(2-0)^2}???

???|a|=\sqrt{9+1+4}???

???|a|=\sqrt{14}???

The magnitude of ???b\langle 1,4,3\rangle??? is

???|b|=\sqrt{(1-0)^2+(4-0)^2+(3-0)^2}???

???|b|=\sqrt{1+16+9}???

???|b|=\sqrt{26}???

Plugging ???a\cdot{b}=5???, ???|a|=\sqrt{14}???, and ???|b|=\sqrt{26}??? into our cosine formula gives

???\cos{\theta}=\frac{5}{\sqrt{14}\sqrt{26}}???

???\cos{\theta}=\frac{5}{\sqrt{364}}???

???\theta=\arccos{\frac{5}{\sqrt{364}}}???

???\theta=74.8^\circ???

The angle between the planes is ???74.8^\circ???.


Want to learn more about Partial Derivatives?

I have a step-by-step course for that. 😃