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, 0≤𝜃≤90∘, then we can write cos𝜃=||⃑𝑛⋅⃑𝑛||‖‖⃑𝑛‖‖‖‖⃑𝑛‖‖, since the scalar product with ⃑𝑛 and ⃑𝑛 is ⃑𝑛⋅⃑𝑛=‖‖⃑𝑛‖‖‖‖⃑𝑛‖‖𝜃‖‖⃑𝑛‖‖‖‖⃑𝑛‖‖𝜃cosorcos and |𝜃|=|𝜃|coscos (because 𝜃=180−𝜃∘).

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, 0≤𝜃≤90∘, and we have cos𝜃=||⃑𝑛⋅⃑𝑛||‖‖⃑𝑛‖‖‖‖⃑𝑛‖‖.

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 −9𝑥−6𝑦+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=||⃑𝑛⋅⃑𝑛||‖‖⃑𝑛‖‖‖‖⃑𝑛‖‖, 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=5√142√57.

To find 𝜃, we use the inverse function on our calculator: 𝜃=5√142√57=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)′′=50.68820581…′′≈51′′.

The angle between the planes is 8648′51′′∘.

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(𝑥−1)+3(𝑦−4)+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=||⃑𝑛⋅⃑𝑛||‖‖⃑𝑛‖‖‖‖⃑𝑛‖‖, 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=|2−6+20|√29√30=16√870, and 𝜃=16√870=57.14936803…≈57,.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 90−𝜃∘ or 𝜃−90∘, 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||⃑𝑑⋅⃑𝑛||‖‖⃑𝑑‖‖‖‖⃑𝑛‖‖ 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=||⃑𝑑⋅⃑𝑛||‖‖⃑𝑑‖‖‖‖⃑𝑛‖‖.

As cossin(90−𝜃)=𝜃∘, it follows that sin𝜃=||⃑𝑑⋅⃑𝑛||‖‖⃑𝑑‖‖‖‖⃑𝑛‖‖.

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=||⃑𝑑⋅⃑𝑛||‖‖⃑𝑑‖‖‖‖⃑𝑛‖‖.

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

However, note that cossin(90−𝜃)=𝜃∘ and sincos(90−𝜃)=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 90−𝜃∘ is the reflection in line 𝑦=𝑥 of point 𝑃 associated with angle 𝜃, meaning that 𝑦=𝑥 and 𝑥=𝑦; therefore, sincos(90−𝜃)=𝜃∘ and cossin(90−𝜃)=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=−7⃑𝑖−⃑𝑗−9⃑𝑘+𝑡2⃑𝑖+⃑𝑗−⃑𝑘 and the plane ⃑𝑟⋅9⃑𝑖−9⃑𝑗+2⃑𝑘=13?

  1. cos7√249498
  2. cos7√4386
  3. sin7√249498
  4. sin7√4386

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=2⃑𝑖+⃑𝑗−⃑𝑘; its components are therefore (2,1,−1).

The vector equation of a plane is of the form ⃑𝑟⋅⃑𝑛=c, where c is a constant. Here, the normal vector of the plane is ⃑𝑛=9⃑𝑖−9⃑𝑗+2⃑𝑘; 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=||⃑𝑑⋅⃑𝑛||‖‖⃑𝑑‖‖‖‖⃑𝑛‖‖.

Let us calculate ||⃑𝑑⋅⃑𝑛||‖‖⃑𝑑‖‖‖‖⃑𝑛‖‖: ||⃑𝑑⋅⃑𝑛||‖‖⃑𝑑‖‖‖‖⃑𝑛‖‖=|2×9+1×(−9)+(−1)×2|2+1+(−1)9+(−9)+2=7√996=72√249=7√249498.

The value 7√249498 is the value of sin𝜃; therefore, 𝜃=7√249498sin. 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=3𝑡−1, 𝑦=−2𝑡+4, 𝑧=5 and the plane 3𝑥−4𝑦+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=||⃑𝑑⋅⃑𝑛||‖‖⃑𝑑‖‖‖‖⃑𝑛‖‖.

Let us calculate ||⃑𝑑⋅⃑𝑛||‖‖⃑𝑑‖‖‖‖⃑𝑛‖‖: ||⃑𝑑⋅⃑𝑛||‖‖⃑𝑑‖‖‖‖⃑𝑛‖‖=|3×3+(−2)×(−4)+0×1|3+(−2)+03+(−4)+1=17√13×26=1713√2.

To find the value of 𝜃, we use on our calculator the inverse function of sine: 𝜃=1713√2𝜃=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)′′=11.51381295…′′≈12′′.

The angle between the planes is 6737′12′′∘.

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 𝑥−77=𝑦−7−5=𝑧−41 and the plane 6𝑥−8𝑦−5𝑧−17=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 𝑥−𝑥𝑙=𝑦−𝑦𝑚=𝑧−𝑧𝑛, 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=||⃑𝑑⋅⃑𝑛||‖‖⃑𝑑‖‖‖‖⃑𝑛‖‖.

Let us calculate ||⃑𝑑⋅⃑𝑛||‖‖⃑𝑑‖‖‖‖⃑𝑛‖‖: ||⃑𝑑⋅⃑𝑛||‖‖⃑𝑑‖‖‖‖⃑𝑛‖‖=|7×6+(−5)×(−8)+1×(−5)|7+(−5)+16+(−8)+(−5)=77√75×125=77√25×3×25×5=7725√15.

To find the value of 𝜃, we use on our calculator the inverse function of sine: 𝜃=7725√15𝜃=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)′′=44.82844578…′′≈45′′.

The angle between the planes is 5240′45′′∘.

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, 0≤𝜃≤90∘, and we have cos𝜃=||⃑𝑛⋅⃑𝑛||‖‖⃑𝑛‖‖‖‖⃑𝑛‖‖.
  • Angle 𝜃 between a plane 𝑃 with normal vector ⃑𝑛 and a line 𝑙 with direction vector ⃑𝑑 is such that sin𝜃=||⃑𝑑⋅⃑𝑛||‖‖⃑𝑑‖‖‖‖⃑𝑛‖‖.

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