- #1
zacman2400
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Homework Statement
Prove that the line of intersection of the planes x+2y-z=2 and 3x+2y+2z=7 is parallel to the line x=1+6t, y=3-5t, z=2-4t. find an equation of the plane determined by the two lines
3. The Attempt at a Solution [/
cross product of n1 by n2 to determine direction of intersecting line, that answer is
(6,-5,-4) which is parallel to l2 as its direction vector is the same so the scalar c=1, thus parallel. From here, I let x=0 in both plane 1 and 2 to determine a position vector for my intersecting line, thus
2y-z=2 and 2y+2z=7
z=2y-2 so y=11/6 and z=10/6, x=0
thus the intersecting line is determined by r=(0,11/6,10/6)+t(6,-5,-4)
now is where I get stuck. Intuitively, I want to find one point p on r say the position vector
(0,11/6,10/6), then find two point p1, p2 on l2, then find two vectors pp1, pp2, cross product them, and find my answer...unfortunately this doesn't work thus after an hour of frustration I beg for help