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

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

Let us eliminate one variable

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

Or

Consider II equation

5x+y-z=4

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

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

Equation of line:-

Parametric equation of line:-

x(t) = 1 - 3t

y(t) = 31t - 1

z(t) = 16t

Step-by-step explanation:

"half of R" it is:  

"y cubed" it is:  

Therefore, half of R subtracted from y cubed is: