Find the coordinates of the point where the line $\displaystyle x = 2 - 3t$, $\displaystyle y = 1 + t$, $\displaystyle z = 4 - t $intersects the plane $\displaystyle 2x - 3y + z = -16$

Please explain the solution with the logic behind it.

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- Nov 2nd 2008, 06:28 AMAltairPoint of intersection of a line with plane
Find the coordinates of the point where the line $\displaystyle x = 2 - 3t$, $\displaystyle y = 1 + t$, $\displaystyle z = 4 - t $intersects the plane $\displaystyle 2x - 3y + z = -16$

Please explain the solution with the logic behind it. - Nov 2nd 2008, 08:16 AMHowardF
Just substiture the parametric values of x y and z into the deifnition of the plane so:-

http://www.mathhelpforum.com/math-he...db259a21-1.gif

becomes

2(2-3t) - 3(1+t) + (4-t) = -16

multiply out and gather terms yields

10t=21

t=2.1

So the line intersects theplane at the parametric value t = 2.1 which defines all the values of x, y and z

(x,y,z) = (-4.3,3.1,1.9)