Find a Cartesian Equation of the plane with contains the line #(x-2)/3=(y+4)/2=(z-1)/2# and passes through the point #(1,1,1)#?

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The answer in the textbook is #-10x-2y+16z=4#

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Answer 1 · 2018-02-17T20:39:37

See below.

Given the line

#L->p=p_0+lambda vec v# with

#p = (x,y,z)#
#p_0 = (2,-4,1)# and
#vec v = (3,2,2)#

and the point

#p_1 = (1,1,1)#

we have the plane equation

#<< p-p_1, vec n >> = 0#

it should obey the condition

#<< p_0-p_1+lambda vec v, vec n >> = 0# or

#<< p_0-p_1, vec n >> + lambda << vec v, vec n >> = 0#

so we conclude

# << p_0-p_1, vec n >> =0#
#<< vec v, vec n >> = 0#

so we conclude that

#(p_0-p_1) bot vec n # and #vec v bot vec n#

so we can choose

#vec n = (p_0-p_1) xx vec v# and then

#Pi -> << p-p_1, (p_0-p_1) xx vec v >> = 0# is the plane equation or putting values

#10x+2y-17z+5=0#