In the above question could anyone please explain me what they have done.
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I am sorry but I do not understand the suggested solution. Anyway, this is an alternative approach.
The cartesian equation of the plane $E$ is $$2x-y-2z+5=0$$ Therefore the distance of $A$ from that plane is $$|AB|=\frac{|2(-4)-(2)-2(2)+5|}{\sqrt{2^2+(-1)^2+(-2)^2}}=3.$$ Moreover $$|AC|=\|(-4-(-3),2-0,2-4)\|=\sqrt{1^2+2^2+(-2)^2}=3.$$ Now note that $AB \perp AC$, hence the required area is $$\frac{|AB|\cdot|AC|}{2}=\frac{9}{2}.$$
Robert Z
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How do you get the equation of a plane – Koolman Dec 19 '16 at 09:14
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And point B also ?? – Koolman Dec 19 '16 at 09:15
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@koolman Let $ax+by+cz+d=0$ the equation of the plane then $(a,b,c)$ is parallel to the cross product of $(1,2,0)$ and $(1,0,1)$. Then find $d$ by imposing that the equation holds for $(-1,1,1)$. – Robert Z Dec 19 '16 at 09:21
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@koolman You need not to find the coordinates of $B$. You need just the distance of $B$ from $A$. – Robert Z Dec 19 '16 at 09:24
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And both AB and AC are perpendicular – Koolman Dec 19 '16 at 09:24
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@koolman $AB$ is orthogonal to the plane $E$ (minimum distance segment) and $AE$ is a parallel to that plane. – Robert Z Dec 19 '16 at 09:29
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Observe that
$$F:\;\;(-4,2,2)+\lambda(1,2,0)+\mu(1,0,1)$$
As $\;E,\,F\;$ are parallel planes, their distance $\;\mathcal D\;$ is the distance between any point in $\;E\;$ to the plane $\;F\;$, so:
$$\mathcal D^2=\min_{\lambda,\mu}d^2\left((-1,1,1)\,,\,\,(\lambda+\mu-4,\,2\lambda+2,\,\mu+2)\right)=$$
$$=\min\left[(\lambda+\mu-3)^2+(2\lambda+1)^2+(\mu+1)^2\right]=9\;\text{(why?)}$$
and thus the minimal distance is $\;3\;$
Since $\;||AB||\;$ is the above minimal distance and $\;C\;$ is on $\;F\;$ , the triangle $\;\Delta ABC\;$ is a straight angle one, with $\;\angle CAB=90^\circ\;$, and thus this triangle's area is simply
$$\frac{AC\cdot AB}2=\frac{3\cdot3}2=\frac92$$
DonAntonio
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