Question #73c53

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Answer 1 · 2018-01-07T12:37:05

See below.

Given

#Pi_1 -> 2x-2y+z+3=0#
#Pi_2 ->2x-2y+z+9 = 0#

Those planes can be represented alternatively as

#Pi_1 -> << p-p_1, vec v >> = 0#
#Pi_2 -> << p-p_2, vec v >> = 0#

then the sought plane can be computed as

#Pi -> << p- 1/2(p_1+p_2), vec v >> = 0# and this justifies

#Pi->2x-2y+z+1/2(3+9) = 2x-2y+z+6 = 0#