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lora16 [44]
3 years ago
6

Where does the line through (1, 0, 1) and (4, −3, 3) intersect the plane x + y + z = 6?

Mathematics
1 answer:
Irina-Kira [14]3 years ago
7 0

Parametrically for real t the line is


(x,y,z) = (1-t)(1,0,1) + t(4,-3,3)


(x,y,z) = (1,0,1) + t(4-1,-3-0,3-1)


(x,y,z) = (1,0,1) + t(3,-3,2)


That's three equations; we substitute into


x+y+z=6


(1+3t)+(0-3t) + ( 1+2t)=6


2t=4


t=2


(x,y,z) = (1,0,1) + 2(3,-3,2) = (7,-6,5)


The intersection is a single point (7,-6,5).



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