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.
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.
Just substiture the parametric values of x y and z into the deifnition of the plane so:-
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)