Question

Find an equation for the plane that (a) is perpendicular to v=(1,1,1) and passes through (1,0,0). (b) is perpendicular to v=(1,2,3) and passes through

Vectors
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asked 2021-05-16
Find an equation for the plane that
(a) is perpendicular to \(v=(1,1,1)\) and passes through (1,0,0).
(b) is perpendicular to \(v=(1,2,3)\) and passes through (1,1,1)
(c) is perpendicular to the line \(l(t)=(5,0,2)t+(3,-1,1)\) and passes through (5,-1,0)
(d) is perpendicular to the line \(l(t)=(-1,-2,3)t+(0,7,1)\) and passes through (2,4,-1).

Answers (1)

2021-05-17
a) \(\vec{n}=(1,1,1)\) where \(\vec{n}\) is the normal to the plane that passes through (1,0,0)
\(\therefore\) Using the equation of the plane we get:
\(1(x-1)+1(y-0)+1(z-0)=0\)
\(x+y+z-1=0\)
b)\(\vec{n}=(1,2,3)\) where \(\vec{n}\) is the normal to the plane that passes through (1,1,1)
\(\therefore\) Using the equation of the plane we get:
\(1(x-1)+2(y-1)+3(z-1)=0\)
\(x+2y+3z-6=0\)
c) \(\vec{n}=\text{direction of line}=(5,0,2)\) where \(\vec{n}\) is the normal to the plane that passes through (5,-1,0)
\(\therefore\) Using the equation of the plane we get:
\(5(x-5)+0(y+1)+2(z-0)=0\)
\(5x+2z-25=0\)
d) \(\vec{n}=\text{direction of line}=(-1,-2,3)\) where \(\vec{n}\) is the normal to the plane that passes through (2,4,-1)
\(\therefore\) Using the equation of the plane we get:
\(-1(x-2)-2(y-4)+3(z+1)=0\)
\(-x-2y+3z+13=0\)
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