Math Problem Statement

Find the equation of L, the line of intersection of planes x−z=1 and y+2z=3. If a plane Q contains the line L and is perpendicular to plane x+y−2z=1, find the equation of plane Q

Solution

To solve the problem step by step:

Step 1: Find the equation of line LL, the intersection of the two planes

The two planes are:

  1. xz=1x - z = 1
  2. y+2z=3y + 2z = 3

Parameterizing the line:

Express x,y,zx, y, z in terms of a parameter tt.

  • From xz=1x - z = 1: x=z+1x = z + 1

  • From y+2z=3y + 2z = 3: y=32zy = 3 - 2z

Let z=tz = t (parameterize zz as tt): x=t+1,y=32t,z=tx = t + 1, \quad y = 3 - 2t, \quad z = t

Thus, the parametric equations of the line LL are: x=t+1,y=32t,z=tx = t + 1, \quad y = 3 - 2t, \quad z = t

Or in vector form: r(t)=(1,3,0)+t(1,2,1)\vec{r}(t) = (1, 3, 0) + t(1, -2, 1)

Step 2: Plane QQ passes through LL and is perpendicular to x+y2z=1x + y - 2z = 1

The plane QQ contains LL, so its normal vector nQ\vec{n}_Q must be perpendicular to the line LL. Also, QQ is perpendicular to the plane x+y2z=1x + y - 2z = 1, so nQ\vec{n}_Q is parallel to the normal vector of this plane, n1=(1,1,2)\vec{n}_1 = (1, 1, -2).

Normal vector to QQ:

The direction vector of LL is dL=(1,2,1)\vec{d}_L = (1, -2, 1). Since QQ contains LL, nQ\vec{n}_Q must be orthogonal to dL\vec{d}_L: nQdL=0\vec{n}_Q \cdot \vec{d}_L = 0

Let nQ=(a,b,c)\vec{n}_Q = (a, b, c). Then: a(1)+b(2)+c(1)=0a(1) + b(-2) + c(1) = 0 a2b+c=0a - 2b + c = 0

Additionally, nQ\vec{n}_Q is parallel to n1=(1,1,2)\vec{n}_1 = (1, 1, -2), so: nQ=k(1,1,2)\vec{n}_Q = k(1, 1, -2)

Let nQ=(1,1,2)\vec{n}_Q = (1, 1, -2) (choosing k=1k = 1).

Equation of plane QQ:

The general equation of a plane is: x+y2z=dx + y - 2z = d

To find dd, substitute a point on LL (e.g., (1,3,0)(1, 3, 0)): 1+32(0)=d1 + 3 - 2(0) = d d=4d = 4

Thus, the equation of plane QQ is: x+y2z=4x + y - 2z = 4


Final Answers:

  1. Equation of line LL: x=t+1,y=32t,z=tx = t + 1, \quad y = 3 - 2t, \quad z = t or: r(t)=(1,3,0)+t(1,2,1)\vec{r}(t) = (1, 3, 0) + t(1, -2, 1)

  2. Equation of plane QQ: x+y2z=4x + y - 2z = 4


Would you like additional clarifications or examples?

Related Questions:

  1. How do you find the intersection of two planes geometrically?
  2. What is the condition for a plane to be perpendicular to another plane?
  3. Can a line lie completely within two planes simultaneously? Why?
  4. How do parameterized line equations relate to vector geometry?
  5. What are some other forms of representing a plane equation?

Tip:

When solving problems involving planes and lines, always check for consistency by substituting values into the equations. This ensures the solution satisfies all given conditions.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Vector Geometry
Equations of Planes
Line-Plane Intersection
Parametric Equations

Formulas

Parametric form of a line: x = x0 + t * dx, y = y0 + t * dy, z = z0 + t * dz
General equation of a plane: ax + by + cz = d
Dot product condition for perpendicular vectors: n1 · n2 = 0

Theorems

Two planes intersect in a line if they are not parallel
A plane perpendicular to another plane has its normal vector parallel to the other's

Suitable Grade Level

Grades 11-12 or College Level