How to find the linear equation of the plane through the point #(1,2,3)# and contains the line represented by the vector equation #r(t)=⟨3t,6−2t,1−2t⟩# ?

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Answer 1 · 2016-06-16T09:37:55

#26 - 4 x + 4 y - 10 z=0#

The parametric line equation is

#p = p_0 + t * vec v#

where #p = {x,y,z}#, #p_0 = {0,6,1}# and #vec v = {3,-2,-2}#

The plane equation can be given by

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

where #p_1 = {1,2,3}# and #vec w# is perpendicular to #vec v# and to the segment #p_1-p_0# so

#vec w = vec v xx (p_1-p_0) = {3,-2,-2} xx{1,-4,-2} #

then

#vec w = {-4, 4, -10}#

The plane equation reads

#(-4)(x-1)+4(y-2)+(-10)(z-3) = 0#

or

#26 - 4 x + 4 y - 10 z=0#