Find an equation for the plane that is perpendicular to the line l(t) = (3, 0, 9)t + (6, −1, 1) and passes through (7, −1, 0).
Mathematics · College · Thu Feb 04 2021
llm
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The line can be written as
(x-6)/3 = (y+1)/0=(z-1)/9= t
The line since perpendicular to the plane is normal to the plane.
Direction ratios of the line are (3,0,9)
SO plane has normal with DR (3,0,9) and plane passes through (7,-1,0)
3(x-7)+0(y+1)+9(z-0) =0
3x+9z =21
x+3z =7 is the equation of the plane.