 # Find an equation for the plane that (a) is perpendicular Joseph Krupa 2021-12-15 Answered
Find an equation for the plane that
(a) is perpendicular to $\text{f{v} = (1, 1, 1}f\left\{v\right\}=\left(1,1,1\right)$ and passes through (1,0,0).
(b) is perpendicular to $\text{f{v} = (1, 2, 3}f\left\{v\right\}=\left(1,2,3\right)$ and passes through (1,1,1).
(c) is perpendicular to the line $\text{f{l}(t) = (5, 0, 2)t + (3, −1, 1}f\left\{l\right\}\left(t\right)=\left(5,0,2\right)t+\left(3,-1,1\right)$ and passes through (5, −1, 0)(5,−1,0).
(d) is perpendicular to the line $\text{f{l}(t) = (−1, −2, 3)t + (0, 7, 1}f\left\{l\right\}\left(t\right)=\left(-1,-2,3\right)t+\left(0,7,1\right)$ and passes through (2, 4, −1)(2,4,−1).
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a) $\stackrel{\to }{n}=\left(1,1,1\right)$ where $\stackrel{\to }{n}$ is the normal to the plane that passes through $\left(1,0,0\right)$
Using the equation of the plane we get:
$1\left(x-1\right)+1\left(y-0\right)+1\left(z-0\right)=0$
$x+y+z-1=0$
b) $\stackrel{\to }{n}=\left(1,2,3\right)$ where $\stackrel{\to }{n}$ is the normal to the plane that passes through (1,1,1)
Using the equation of the plane we get
$1\left(x-1\right)+2\left(y-1\right)+3\left(z-1\right)=0$
$x+2y+3z-6=0$

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c) $\stackrel{\to }{n}=$ direction of line $=\left(5,0,2\right)$ where $\stackrel{\to }{n}$ is the normal to the plane that passes through $\left(5,-1,0\right)$
Using the equation of the plane we get:
$5\left(x-5\right)+0\left(y+1\right)+2\left(z-0\right)=0$
$5x+2z-25=0$
d) $\stackrel{\to }{n}=$ direction of line $=\left(-1,-2,3\right)$ where $\stackrel{\to }{n}$ is the normal to the plane that passes through (2,3,-1)
Using the equation of the plane we get:
$-1\left(x-2\right)-2\left(y-4\right)+3\left(z+1\right)=0$
$-x-2y+3z+13=0$

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