Intersection of Planes Cases of Intersection - Two planes in R3 Using two sheets of paper to model two planes, answer the following questions. How can the system have exactly one solution? How can system have more than one solution? How can the system have no solutions? Are there any other cases? Summary: In R3 , The three cases for the intersection of two planes are: i) Intersect in a line - Occurs when the two planes are not parallel. ii) Intersect never - Occurs when the two planes are parallel, but distinct. iii) Intersect everywhere - Occurs when the two planes are parallel and coincident. i)

ii)

iii)

The easiest way to tell if two planes are parallel is to compare their normal vectors, which you can get easily from the scalar equation. Example 1.1. Will the planes P1 : 2x − y + z − 1 = 0 and P2 : x + y + z − 6 = 0 intersect? If so, find the equation for the line of the intersection. The normal vector for P1 is n~1 = [2, −1, 1] and the normal vector for P2 is n~2 = [1, 1, 1]. By inspection these are not scalar multiples of each other and not parallel, therefore P1 and P2 must intersect in a line. We can find the line of intersection by solving the system of equations 2x + y + z − 1

=

0

(Equation 1)

x+y+z−6

=

0

(Equation 2)

If we subtract equation 2 from equation 1, then we can solve for an x value. 2x + y + z − 1 − (x + y + z − 6) = 0 − 0 ⇒ 2x − x + y − y + z − z − 1 + 6 = 0 ⇒x+5=0 ⇒ x = −5. Substituting x = −5 into equation 1, we get 2(−5) + y + z − 1 = 0 ⇒ y = 6 − z. 1

Since we are defining a line in R3 we must either use parametric equations or a vector equation, so we’ll need to introduce the parameter t somehow. After solving for x and y we had the z variable left over. This leftover is the one that is set equal to t in order to make the equation of a line. It is not always z that is left over either. If you solve for x and z first then y will be the left over variable and it will be set equal to t. Here are both the parametric and vector equations for the intersection `. Parametric Equations

Vector Equation

x=0

` : [x, y, z] = [0, 6, 0] + t[0, −1, 1].

y =6−t z = t. Example 1.2. Will the planes P1 : 5x − 3y − 2z = 7 and P2 : 2x − y + 3z = 4 intersect? If so, find the equation for the line of the intersection. The normal vector for P1 is n~1 = [5, −3, −2] and the normal vector for P2 is n~2 = [2, −1, 3]. By inspection these are not scalar multiples of each other and not parallel, therefore P1 and P2 must intersect in a line. We can find the line of intersection by solving the system of equations 5x − 3y − 2z

=

7

(Equation 1)

2x − y + 3z

=

4

(Equation 2)

If we subtract 3 times equation 2 from equation 1, then we can solve for an expression for x. 5x − 3y − 2z − 1 − 3(2x − y + 3z) = 7 − 3(4) ⇒ 5x − 6x − 3y + 3y − 2z − 3z = 7 − 12 ⇒ −x − 5z = −5 ⇒ x = −5z + 5. Substituting x = −5z + 5 into equation 2, we get 2(−5z + 5) − y + 3z = 4 ⇒ −10z + 10 − y + 3z = 4 ⇒ y = −7z + 6 Setting z = t we get the following parametric and vector equations for the intersection `. Parametric Equations

Vector Equation

x = 5 − 5t

` : [x, y, z] = [5, 6, 0] + t[−5, −7, 1].

y = 6 − 7t z = t.

2

Example 1.3. Will the planes P1 : x + y − 2z + 2 = 0 and P2 : 2x + 2y − 4z + 4 = 0 intersect? If so, find the equation for the line of the intersection. Since the normal for P1 is n~1 = [1, 1, −2] and the normal for P2 is n~2 = [2, 2, −4], P1 and P2 are parallel since 2n~1 = n~2 . By the summary above, either P1 and P2 intersect everywhere or nowhere. We’ll find a point on P1 and if that point also lies on P2 we’ll know that the two planes are coincident. The plane P1 has an z-intercept at [0, 0, 1], and when we test that in P2 , 2(0) + 2(0) − 4(1) + 4 = 0 so the we can conclude that P1 and P2 are coincident. *The two planes satisfy the equation 2P1 = P2 , which is another way to check if the two planes are parallel and coincident. If P1 6= cP2 , but the normal vectors of the two planes are parallel, then the planes are parallel and distinct. Cases of Intersection - Three Planes in R3 Using three sheets of paper to model three planes, consider how many different intersection cases you can create by: adding a third plane to case i) for two planes, adding a third plane to case ii) for two planes, adding a third plane to case iii) for two planes, or being creative. Summary: In R3 , The six cases for the intersection of three planes are: i) Intersect in a point - Occurs when 3 planes intersect and create ‘corners’, similar to the corners of the 8 octants of R3 . ii) Intersect in a line - Occurs when 3 planes intersect, making a shape resembling a fan or a paddle wheel. iii) Intersect everywhere - Occurs when 3 planes are parallel and coincident. iv) Intersect never - Occurs when all 3 planes are parallel and distinct. v) Intersect in two lines - Occurs when two parallel and distinct planes are met by a third nonparallel plane, resembling a 3D ‘Z’. vi) Intersect in three lines - Occurs when 3 planes intersect and form a triangular tunnel. i)

ii)

iii)

iv)

v)

vi)

3

The first three cases are called consistent systems of 3 planes, meaning that all three planes intersect with a common solution. The last three cases are inconsistent systems of 3 planes, having no common intersection point for all three planes. Definition 1.4. A set of points or system of lines are coplanar if the they all lie on a common plane. This is the higher dimensional version of the term collinear. If a set of points are collinear, then they are coplanar as well.

For vectors ~a, ~b, and ~c, the three are coplanar if ~a · ~b × ~c = 0. In other words, ~b × ~c will be perpendicular to ~a if ~a, ~b, and ~c are coplanar. The diagram on the left shows such an arrangement, and ~b × ~c is indeed perpendicular to ~a.

When analyzing the cases of 6 planes intersecting, it is not as simple as deciding if they are parallel or non-parallel. We’ll need some specific information about the normals of each plane. i) The normals are not coplanar. ii) The normals are coplanar, but not parallel. iii) The normals are parallel (and of course coplanar). iv) The normals are parallel. v) Two of the normals are parallel and the other is not. vi) The normals are coplanar but not parallel. Case ii) and vi), as well as case iii) and case iv) have the same conditions on normal vectors, so how are they different? In each situation, the difference will be if the system of 3 planes is consistent or not. Suppose the normals for a system of 3 planes are coplanar but not parallel. Then, either the planes are arranged in case ii) or vi). If you solve the system of equations, and you get a solution, then the planes resemble case ii). If there is no solution, then the planes resemble case vi). Example 1.5. Determine how the following system of 3 planes intersect, and state the solution if it exists. P1 : 2x − y + 4z = 6 P2 : 2x − y + 4z = 8 P3 : 4x − 2y + 8z = 0 Normal Analysis- The normal vectors or all 3 planes are parallel, therefore the planes intersect as in either case iii) or case iv). We may skip the coplanar test since parallel vectors are also coplanar. To further distinguish between the two cases, we must solve for a solution. 4

Solving the System- Rather than solve the system of equations, we can implement the fact that the scalar equations of parallel planes will be scalar multiples of each other if they are coincident. By inspection, these three are not scalar multiples of each other. Therefore, the system is an inconsistent system made up of 3 distinct parallel planes. Example 1.6. Determine how the following system of 3 planes intersect, and state the solution if it exists. P1 : 2x − y + 5z = 7 P2 : 3x − 23 y +

15 2 z

=8

P3 : x + 2y + 4z = 7 Normal Analysis- the normals for P1 and P2 are actually parallel, n~2 =

3 ~1 . 2n

The third normal

however is not parallel. This information strongly suggests that the planes intersect as in case v) (so we will skip the coplanar test), but we must be careful. If P1 and P2 are coincident the planes will actually intersect in a line, as a special version of case ii). However, since 8 6=

3 2 (7),

P1 and P2 are not be coincident. Therefore the system of 3 planes is

inconsistent and intersect only in two pairs, forming two lines. Example 1.7. Determine how the following system of 3 planes intersect, and state the solution if it exists. P1 : x − y + 2z = 3 P2 : 2x + 3y − z = 1 P3 : x − 4y + 3z = 22 Normal Analysis- The three normal vectors are n~1 = [1, −1, 2], n~2 = [2, 3, −1], and n~3 = [1, −4, 3]. It’s easy to prove that none of these vectors are parallel, so the next we must see if they are collinear. n~2 × n~3 = [(3)(3) − (−1)(−4), (−1)(1) − (2)(3), (2)(−4) − (3)(1)] = [5, −6, −11] ∴ n~1 · n~2 × n~3 = [1, −1, 2] · [5, −6, −11] ⇒ n~1 · n~2 × n~3 = 5 + 6 − 22 6= 0, which means that the normals are not coplanar. This corresponds only to case i), which means that the system of 3 planes is consistent and they intersect in a point. Solving the System- With three equations, our method changes. We create two pairs of equations and solve for their respective intersections first. Let’s make the following pairs: P1 , P2 ; P1 , P3 and set up the system of equations. x − y + 2z = 3

x − y + 2z = 3

2x + 3y − z = 1

x − 4y + 3z = 22

The quickest way to solve this system is to eliminate the same variable in each case. We’ll eliminate x in both cases.

5

P2 − 2P1

P1 − P2

2x + 3y − z − 2(x − y + 2z) = 1 − 2(3)

x − y + 2z − (x − 4y + 3z) = 3 − 22

⇒ 5y − 5z = −5

⇒ 3y − z = −25

⇒ y = z − 1.

⇒y=

z 3

−

25 3 .

Next, we take the resulting equations and find their intersection y =z−1 y=

z 3

−

25 3

⇒ 32 z = 1 −

25 3

⇒ z = −11. We can substitute this value to get the rest of the coordinates. y = (−11) − 1 = −12

x − (−12) + 2(−11) = 3 ⇒ x = 13.

Therefore the three planes intersect at the point (13, −12, −11). Exercise: Suppose three planes P1 , P2 and P3 have non-parallel but coplanar normals. How would you tell if there is an intersection point?

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