Respuesta :
Answer:
Step-by-step 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
[tex]x-3y+6z=4 \\5x+y-z=4[/tex]
Let us eliminate one variable
Multiply I equation by 5 and subtract II from 5 * I
[tex]5x-15y+30z=20 \\5x+y-z=4[/tex]
[tex]-16y+31z=16 \\[/tex]
[tex]16y+16 = 31z[/tex]
Or [tex]y+1 =\frac{31z}{16}[/tex]
Consider II equation
5x+y-z=4
[tex]5x+\frac{31z}{16}-1 -z=4\\5x+\frac{15z}{16}=5\\x=\frac{3z-16}{16}[/tex]
Let z=z
x = \frac{3z-16}{16} and y = \frac{31z}{16}-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, [tex]n_1=<1,-3,6>[/tex]
- Equation of 2nd plane: 5x+y-z=4
- Normal vector of this plane, [tex]n_1=<5,1,-1>[/tex]
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 [tex]n_1\times n_2[/tex]
[tex]\bec{b}=<1,-3,6>\times<5,1,-1>[/tex]
[tex]\vec{b}=<-3,31,16>[/tex]
Equation of line:-
[tex]\dfrac{x-1}{-3}=\dfrac{y+1}{31}=\dfrac{z-0}{16}=t[/tex]
Parametric equation of line:-
x(t) = 1 - 3t
y(t) = 31t - 1
z(t) = 16t
