# SOLUTION: Hi again, hopefully someone can help me with the follwoing question, Find an equation of the line intersection of the two planes: x+2y-z = -1 and 3x-2y+5z = 5 Any help would be

Algebra ->  Algebra  -> Vectors -> SOLUTION: Hi again, hopefully someone can help me with the follwoing question, Find an equation of the line intersection of the two planes: x+2y-z = -1 and 3x-2y+5z = 5 Any help would be       Log On

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 Click here to see ALL problems on Vectors Question 206224: Hi again, hopefully someone can help me with the follwoing question, Find an equation of the line intersection of the two planes: x+2y-z = -1 and 3x-2y+5z = 5 Any help would be most appreciated. -nickAnswer by Alan3354(30993)   (Show Source): You can put this solution on YOUR website!Find an equation of the line intersection of the two planes: x+2y-z = -1 and 3x-2y+5z = 5 --------------- The vector v1 = (1,2,-1) is normal to x+2y-z = -1. The vector v2 = (3,-2,5) is normal to 3x-2y+5z = 5. These vectors are not parallel, so the planes do intersect. ----------------- The cross product gives a vector that's perpendicular to both above. v3 = v1Xv2 = i - j - k (Cross product of v1 and v2), or (1,-1,-1) v3 gives the direction of the line of intersection. -------------- Next, we need a point that is on both planes. +x+2y-z = -1 3x-2y+5z = 5 -------------- Set one dimension to zero, and get 2 linear eqns: x=0 2y - z = -1 2y -5z = -5 ----------- Subtract 2 from 1 4z = 4 z = 1 y = 0 ----- So the point with the position vector (0,0,1) lies on the line of intersection. The eqn of the line of intersection is (0,0,1) + t(1,-1,-1) ----------------- To check a few points: +x+2y-z = -1 3x-2y+5z = 5 t = 1 --> (1,-1,0) is on both planes. t = 4 --> (4,-4,-3) is also on both planes. 2 points is sufficient to fix a line, so it's correct.