The three planes have normals that are coplanar and the planes intersect in a line.

The line of intersection of the planes intersects with another line [x,y,z]=[14,-13,4]+t[2,-1,3]. Find the point of intersection of the two lines.

2x+y+z-7=0

4x+3y-3z-13=0

4x+2y+2z-14=0

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We have the linear system

2x + y + z = 7

4x + 3y - 3z = 13

4x + 2y + 2z = 14

Writing in shortand we have

`[[2,1,1],[4,3,-3],[4,2,2]]|[[7],[13],[14]]`

Using linear row-operations to achieve echelon form, successively we get

`[[1,1/2,1/2],[4,3,-3],[4,2,2]]|[[7/2],[13],[14]]` , `[[1,1/2,1/2],[0,1,-5],[0,0,0]]|[[7/2],[-1],[0]]` , `[[1,0,3],[0,1,-5],[0,0,0]]|[[4],[-1],[0]]`

The planes then intersect on the line given by

`(x,y,z) = (4,-1,0) + t(-3,5,1)`

To find where this line intersects the given line, we equate them and solve for t:

`([4],[-1],[0]) + t([-3],[5],[1]) = ([14],[-13],[4]) + t([2],[-1],[3])`

Rearranging this gives

`t([5],[-6],[2]) = ([-10],[12],[-4])`

which implies that t = z = -2, which further implies that x = 10 and y = -11

**The two lines intersect at (x,y,z) = (10,-11,-2)**

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