Find the point of intersections of line passing through $(0, 0, 1)$ and intersecting line $x + 2 y + z = 1, - x + y - 2 z = 2$ and $x + y = 2, x + z = 2$ with the $x y$ plane.

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Solution

Equation of the line is
$\begin{matrix} x + 2 y + z - 1 + λ (- x + y - 2 z - 2) = 0 x + y - z + μ (x + z - 2) = 0 T h i s l i n e p a s s e s t h r o u g h t h e p o i n t (0, 0, 1) λ = 0, μ = - 2 x + 2 y + z - 1 = 0 = - x + y - 2 z + 2 F o r t h e p o i n t o f i n t e r sec t i o n w i t h p l a n e p u t z = 0 x + 2 y = 1 - x + y + 2 = 0 (\frac{5}{3}, - \frac{1}{3}, 0) \end{matrix}$

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Find the point of intersections of line passing through (0,0,1) and intersecting line x+2y+z=1,−x+y−2z=2 and x+y=2,x+z=2 with the xy plane.

Equation of the line is x+2y+z−1+λ(−x+y−2z−2)=0x+y−z+μ(x+z−2)=0Thislinepassesthroughthepoint(0,0,1)λ=0,μ=−2x+2y+z−1=0=−x+y−2z+2Forthepointofintersectionwithplaneputz=0x+2y=1−x+y+2=0(53,−13,0)

Find the point of intersections of line passing through $(0, 0, 1)$ and intersecting line $x + 2 y + z = 1, - x + y - 2 z = 2$ and $x + y = 2, x + z = 2$ with the $x y$ plane.