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, $n_1=$

• Equation of 2nd plane: 5x+y-z=4

• Normal vector of this plane, $n_1=$

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 $n_1\times n_2$

$\bec{b}=\times$

$\vec{b}=$

Equation of line:-

$\dfrac{x-1}{-3}=\dfrac{y+1}{31}=\dfrac{z-0}{16}=t$

Parametric equation of line:-

x(t) = 1 - 3t

y(t) = 31t - 1

z(t) = 16t

Answer 2

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

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

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 $y+1 =\frac{31z}{16}$

Consider II equation

5x+y-z=4

$5x+\frac{31z}{16}-1 -z=4\\5x+\frac{15z}{16}=5\\x=\frac{3z-16}{16}$

Let z=z

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