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# Find an equation of the line of intersection of the planes Q and R. Q: x - y - 2z = 1; R: x + y + z = -1

**Solution:**

The equation of plane Q is x - y - 2z = 1

The equation of Plane R is x + y + z = -1

The line of intersection of the two planes is perpendicular to both the planes’ normal vectors n1 and n1 and therefore parallel to n1 × n1 is a vector parallel to the planes’ line of intersection. In our case,

The vector v = i - 3j + 2k is parallel to the line of intersection of the planes.

To find a point on the line, we can take any point common to the two planes.

Substituting z = 0 in the plane equations and solving simultaneously identifies one of the points as (0, -1, 0) as shown below

x - y = 1

x + y = -1

2x = 0

x=0

y=-1

The line is

x = 0 + t; y = -1 - 3t; z = 0 + 2t

## Find an equation of the line of intersection of the planes Q and R. Q: x - y - 2z = 1; R: x + y + z =-1

**Summary:**

The equation of the line of intersection of the planes Q and R. Q: x - y - 2z = 1; R: x + y + z =-1 is x = 4t; y = -1 + 2t; z = 3t

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