Question

Find the intersection of the two planes with equations 5(x − 1) + 2y + (z + 1) = 0 and (x − 1) + 7y − (z + 1) = 0. (use the parameter t. enter your answers as a comma-separated list of equations.)

Answer 1

The intersection of two planes can be found by solving the system of equations formed by the equations of the planes. In this case, the intersection is given by the equations x = (4 - 9y) / 6, y = y, and z = -3 - (23y/6), where y can be any real number.

Step-by-step explanation:

The intersection of two planes can be found by solving the system of equations formed by the equations of the planes.

To find the intersection, we can set the two equations equal to each other and solve for x, y, and z.

1. Rewrite the equations in standard form: 5x - 5 + 2y + z + 1 = 0 and x - 1 + 7y - z - 1 = 0
2. Add the equations together: 6x + 9y = 4
3. Solve for x in terms of y: x = (4 - 9y) / 6
4. Substitute the value of x into one of the original equations to solve for z: 5((4 - 9y) / 6 - 1) + 2y + (z + 1) = 0
5. Simplify and solve for z: z = -3 - (23y/6)

Therefore, the intersection of the two planes is given by the equations x = (4 - 9y) / 6, y = y, and z = -3 - (23y/6), where y can be any real number.

Answer 2

Given:
P1: 5(x-1)+2y+1(z+1)=0
P2: 1(x-1)+7y-1(z+1)=0
Need the intersection of the planes P1 and P2 (a line)

By inspection of the equations,
normal to P1: N1<5,2,1>
normal to P2: N2<1,7,-1>
Direction vector, V, of the required line is the cross product of P1 & P2:
i j k
5 2 1
1 7 -1
=V<-9, 6, 33>

Since P1 passes through point (1,0,-1), the parametric equation of the required line is <1-9t, 0+6t, -1+33t>

L: