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(1 point) The planes 3x + 4y + z = 2 and 3x – 3y = -18 are not parallel, so they must intersect along a line that is common t

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Answer #1

Given that the planes 3x+4y+z=2 and 3x-3y=-18 are not parallel , so they must intersect along a line that is common to both of them .

For finding the intersecting line now we solve the equations ,

3x+4y+z=2 and 3x-3y=-18

These are two equations of three variables so we have a free variable .Let z = t then we get ,

3x+4y=2-t and 3x-3y=-18

Solving these equations we get , 7y=20 - t and 7x= -22 - t

That means , y = (20 - t)/7 and x = (-22 - t)/7

Therefore (x,y,z)=((-22 - t)/7 , (20 - t)/7, t )

=(-22/7 , 20/7 , 0) + t*(-1/7 , -1/7 ,1)

Thus the vector parametric equation for the intersecting line is

L(t) = (-22/7 , 20/7 , 0) + t*(-1/7 , -1/7 , 1)

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