Question

Find a set of parametric equations for the line of intersection of the planes x-3y+6z=4 5x+y-z=4

Answer 1

Parametric equation of line of intersections of the planes.

x(t) = 1 - 3t

y(t) = 31t - 1

z(t) = 16t

Explanation:

The equation of planes are x-3y+6z=4 and 5x+y-z=4

• Equation of first plane: x-3y+6z=4
• Normal vector of this plane,
  `n_1=<1,-3,6>`

• Equation of 2nd plane: 5x+y-z=4

• Normal vector of this plane,
  `n_1=<5,1,-1>`

Point of intersection of x-3y+6z=4 and 5x+y-z=4

Let z = 0

x - 3y = 4 and 5x + y = 4

solve for x and y

x = 1 and y = -1

Point of intersection of plane, (1,-1,0)

The line of intersection of both plane must be passes through this point.

Parallel vector of line is cross product of
`n_1* n_2`

`\bec{b}=<1,-3,6>*<5,1,-1>`

`\vec{b}=<-3,31,16>`

Equation of line:-

`(x-1)/(-3)=(y+1)/(31)=(z-0)/(16)=t`

Parametric equation of line:-

x(t) = 1 - 3t

y(t) = 31t - 1

z(t) = 16t

Answer 2

Explanation:

Two equations of planes are given.

We have to find the equation of the line which is formed by intersection of these two planes in parametric form

`x-3y+6z=4 \\5x+y-z=4`

Let us eliminate one variable

Multiply I equation by 5 and subtract II from 5 * I

`5x-15y+30z=20 \\5x+y-z=4`

`-16y+31z=16 \\`

`16y+16 = 31z`

Or
`y+1 =(31z)/(16)`

Consider II equation

5x+y-z=4

`5x+(31z)/(16)-1 -z=4\\5x+(15z)/(16)=5\\x=(3z-16)/(16)`

Let z=z

x = \frac{3z-16}{16} and y = \frac{31z}{16}-1