 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 Armorikam 2021-05-16 Answered
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).

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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:
$$-1(x-2)-2(y-4)+3(z+1)=0$$
$$-x-2y+3z+13=0$$