Lesson Explainer: The Angle between Two Planes and between a Plane and a Straight Line Mathematics

In this explainer, we will learn how to find the measure of an angle between two planes or that between a line and a plane.

Let us consider two intersecting planes 𝑃 and 𝑄.

The angle between them can be visualized in a plane perpendicular to their line of intersection. We see that, as for two intersecting lines, there are two angles between them, 𝜃 and 𝜃, with 𝜃+𝜃=180.

Drawing now the two pairs of possible normal vectors of the two planes, we see that either 𝜃 or 𝜃 is the angle between the normal vectors, depending on the directions of the normal vectors.

If we define the angle 𝜃 between two planes 𝑃 and 𝑄 as the acute angle between them, that is, 0𝜃90, then we can write cos𝜃=||𝑛𝑛||𝑛𝑛, since the scalar product with 𝑛 and 𝑛 is 𝑛𝑛=𝑛𝑛𝜃𝑛𝑛𝜃cosorcos and |𝜃|=|𝜃|coscos (because 𝜃=180𝜃).

Definition: Angle between Two Planes

The angle 𝜃 between two planes 𝑃 and 𝑄 with normal vectors 𝑛 and 𝑛 is defined as the acute angle between them; hence, 0𝜃90, and we have cos𝜃=||𝑛𝑛||𝑛𝑛.

Let us look at the first example.

Example 1: Finding the Angle between Two Planes given Their General Equations

Find, to the nearest second, the measure of the angle between the planes 9𝑥6𝑦+5𝑧=8 and 2𝑥+2𝑦+7𝑧=8.

Answer

To find the angle between the planes, we need the normal vectors of the two planes. Knowing that the general equations of a plane are of the form 𝑎𝑥+𝑏𝑦+𝑐𝑧=𝑑, with (𝑎,𝑏,𝑐) being the components of a normal vector of the plane, we find that the components of the normal vectors of the two planes are (9,6,5) and (2,2,7). The angle 𝜃 between the two planes is such that cos𝜃=||𝑛𝑛||𝑛𝑛, where 𝑛 and 𝑛 are two normal vectors of the two planes.

We have cos𝜃=|9×2+(6)×2+5×7|81+36+25×4+4+49=514257.

To find 𝜃, we use the inverse function on our calculator: 𝜃=514257=86.81408006.cos

We need to give the angle to the nearest second. Remember that one degree is 60 minutes (1=60) and one minute is 60 seconds (1=60).

Therefore, 0.81408006=(0.81408006×60)=48.84480343, and 0.84480343=(0.84480343×60)=50.6882058151.

The angle between the planes is 864851.

Let us now find the angle between two planes when the equation of one of the planes is a general equation and the other is a vector equation.

Example 2: Finding the Angle between Two Planes given Their Standard and Vector Equations

Find, to the nearest degree, the measure of the angle between the planes 2(𝑥1)+3(𝑦4)+4(𝑧+5)=0 and 𝑟(1,2,5)=16.

Answer

To find the angle between the planes, we need the normal vectors of the two planes. The coefficients of 𝑥, 𝑦, and 𝑧 in the general equation are the 𝑥-, 𝑦-, and 𝑧-components of a normal vector of the plane. In the equation given, we see that expanding the brackets will each time give a term in 𝑥, 𝑦, or 𝑧 and a constant. The coefficients of 𝑥, 𝑦, and 𝑧 are therefore given by the coefficient in front of each bracket. So, a normal vector of the first plane is (2,3,4).

The other plane equation is given in vector form, where (1,2,5) is a normal vector of the plane.

The angle 𝜃 between the two planes is such that cos𝜃=||𝑛𝑛||𝑛𝑛, where 𝑛 and 𝑛 are two normal vectors of the two planes. Hence, cos𝜃=|(2,3,4)(1,2,5)|2+3+41+(2)+5=|26+20|2930=16870, and 𝜃=16870=57.1493680357,.costothenearestwholenumber

The angle between the two planes to the nearest degree is 57.

Consider now a line 𝑙 intersecting a plane 𝑃.

The angle between line 𝑙 and plane 𝑃 is defined as the smallest possible angle between line 𝑙 and any line in plane 𝑃 that intersects 𝑙 (i.e., that goes through the intersection point of 𝑙 with 𝑃). Hence, it is the acute angle between line 𝑙 and intersection line 𝑙 between plane 𝑃 and plane 𝑁, which is the plane perpendicular to 𝑃 that contains 𝑙. It is therefore defined as the complementary angle to the smallest possible angle between line 𝑙 and the normal of plane 𝑃, as we can visualize in plane 𝑁 shown in the next figure.

The direction vector 𝑑 of 𝑙 and the normal vector 𝑛 of 𝑃 are also represented. The vector indicated with a dashed line is a normal vector of 𝑃 with an opposite direction to that of 𝑛.

We find that the angle 𝜃 between line 𝑙 and plane 𝑃 is either 90𝜃 or 𝜃90, where 𝜃 and 𝜃 are the two possible angles between 𝑑 and 𝑛, depending on their respective directions. (Note that we would find angle 𝜃 between 𝑛 as shown here and the direction vector of 𝑙 in the opposite direction to 𝑑 as shown in the diagram.) However, as for the angle between two planes, we know that cos||𝑑𝑛||𝑑𝑛 will give an acute angle (here 𝜃), since cos𝜃 and cos𝜃 have the same absolute value but cos𝜃 is negative while cos𝜃 is positive.

Therefore, angle 𝜃 between line 𝑙 and plane 𝑃 is complementary to 𝜃(𝜃+𝜃=90), with cos𝜃=||𝑑𝑛||𝑑𝑛.

As cossin(90𝜃)=𝜃, it follows that sin𝜃=||𝑑𝑛||𝑑𝑛.

Definition: Angle between a Line and a Plane

Angle 𝜃 between a plane 𝑃 with normal vector 𝑛 and a line 𝑙 with direction vector 𝑑 is defined as the complementary angle to the smallest possible angle between line 𝑙 and the normal of the plane 𝑃.

It is such that sin𝜃=||𝑑𝑛||𝑑𝑛.

We can easily see in a right triangle why if 𝜃+𝜃=90, then sincos𝜃=𝜃.

However, note that cossin(90𝜃)=𝜃 and sincos(90𝜃)=𝜃 are true equalities beyond the case of right triangles (i.e., also for angles 90) because of the symmetries of the sine and cosine functions, as illustrated here with unit circles. Point 𝑃 associated with angle 90𝜃 is the reflection in line 𝑦=𝑥 of point 𝑃 associated with angle 𝜃, meaning that 𝑦=𝑥 and 𝑥=𝑦; therefore, sincos(90𝜃)=𝜃 and cossin(90𝜃)=𝜃.

Let us see in the next example how to find the angle between a plane and a line given their vector equations.

Example 3: Finding the Angle between a Plane and a Line given their Vector Equations

Which of the following is the smaller angle between the straight line 𝑟=7𝑖𝑗9𝑘+𝑡2𝑖+𝑗𝑘 and the plane 𝑟9𝑖9𝑗+2𝑘=13?

  1. cos7249498
  2. cos74386
  3. sin7249498
  4. sin74386

Answer

To find the angle between a line and a plane, we need to know the components of a direction vector of the line and of a normal vector of the plane. In a vector equation of a line, the direction vector is the vector that is multiplied by 𝑡. Here, it is 𝑑=2𝑖+𝑗𝑘; its components are therefore (2,1,1).

The vector equation of a plane is of the form 𝑟𝑛=c, where c is a constant. Here, the normal vector of the plane is 𝑛=9𝑖9𝑗+2𝑘; its components are (9,9,2).

The angle 𝜃 between a plane 𝑃 with normal vector 𝑛 and a line 𝑙 with direction vector 𝑑 is such that sin𝜃=||𝑑𝑛||𝑑𝑛.

Let us calculate ||𝑑𝑛||𝑑𝑛: ||𝑑𝑛||𝑑𝑛=|2×9+1×(9)+(1)×2|2+1+(1)9+(9)+2=7996=72249=7249498.

The value 7249498 is the value of sin𝜃; therefore, 𝜃=7249498sin. The correct answer is C.

Let us apply the same idea for the last two examples but with different types of line equations.

Example 4: Finding the Angle between a Plane and a Line given their General and Parametric Equations

Find, to the nearest second, the measure of the angle between the straight line 𝑥=3𝑡1, 𝑦=2𝑡+4, 𝑧=5 and the plane 3𝑥4𝑦+𝑧=2.

Answer

To find the angle between a line and a plane, we need to know the components of a direction vector of the line and of a normal vector of the plane. In the parametric equations of a line, the 𝑥-, 𝑦-, and 𝑧-components of the direction vector are the coefficient of 𝑡 in the equation for 𝑥, 𝑦, and 𝑧 respectively. We find that the components of the direction vector of the given line are (3,2,0).

The general equation of a plane is of the form 𝑎𝑥+𝑏𝑦+𝑐𝑧=𝑑, with (𝑎,𝑏,𝑐) being the components of a normal vector of the plane. Hence, the components of the normal vector of the plane here are (3,4,1).

The angle 𝜃 between a plane 𝑃 with normal vector 𝑛 and a line 𝑙 with direction vector 𝑑 is such that sin𝜃=||𝑑𝑛||𝑑𝑛.

Let us calculate ||𝑑𝑛||𝑑𝑛: ||𝑑𝑛||𝑑𝑛=|3×3+(2)×(4)+0×1|3+(2)+03+(4)+1=1713×26=17132.

To find the value of 𝜃, we use on our calculator the inverse function of sine: 𝜃=17132𝜃=67.61986495.sin

We need to give the angle to the nearest second. Remember that one degree is 60 minutes (1=60) and one minute is 60 seconds (1=60).

Therefore, 0.61986495=(0.61986495×60)=37.19189688, and 0.19189688=(0.19189688×60)=11.5138129512.

The angle between the planes is 673712.

Example 5: Finding the Angle between a Plane and a Line given their General and Cartesian Equations

Find, to the nearest second, the measure of the smaller angle between the straight line 𝑥77=𝑦75=𝑧41 and the plane 6𝑥8𝑦5𝑧17=0.

Answer

To find the angle between a line and a plane, we need to know the components of a direction vector of the line and of a normal vector of the plane. The Cartesian equation of a line is of the form 𝑥𝑥𝑙=𝑦𝑦𝑚=𝑧𝑧𝑛, where (𝑥,𝑦,𝑧) are the coordinates of a point that lies on the line and (𝑙,𝑚,𝑛) is a direction vector of the line. Hence, the components of the direction vector of the given line are (7,5,1).

The general equation of a plane is of the form 𝑎𝑥+𝑏𝑦+𝑐𝑧=𝑑, with (𝑎,𝑏,𝑐) being the components of a normal vector of the plane. Hence, the components of the normal vector of the plane here are (6,8,5).

Angle 𝜃 between a plane 𝑃 with normal vector 𝑛 and a line 𝑙 with direction vector 𝑑 is such that sin𝜃=||𝑑𝑛||𝑑𝑛.

Let us calculate ||𝑑𝑛||𝑑𝑛: ||𝑑𝑛||𝑑𝑛=|7×6+(5)×(8)+1×(5)|7+(5)+16+(8)+(5)=7775×125=7725×3×25×5=772515.

To find the value of 𝜃, we use on our calculator the inverse function of sine: 𝜃=772515𝜃=52.67911901.sin

We need to give the angle to the nearest second. Remember that one degree is 60 minutes (1=60) and one minute is 60 second (1=60).

Therefore, 0.67911901=(0.67911901×60)=40.74714076, and 0.74714076=(0.74714076×60)=44.8284457845.

The angle between the planes is 524045.

Key Points

  • Angle 𝜃 between two planes 𝑃 and 𝑄 with normal vectors 𝑛 and 𝑛 is defined as the acute angle between them; hence, 0𝜃90, and we have cos𝜃=||𝑛𝑛||𝑛𝑛.
  • Angle 𝜃 between a plane 𝑃 with normal vector 𝑛 and a line 𝑙 with direction vector 𝑑 is such that sin𝜃=||𝑑𝑛||𝑑𝑛.

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