# Normal Form Planes

Having a hard time solving this without a perpendicular line, would appreciate help.

Find an equation in normal form $ax+by+cz=d$ for the plane P that passes through point $(1,0,-1)$ and contains the line $(x,y,z)=(1-t,t,2)$

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Well, it must be that

$$a-c=d\;\;\;,\;\;\;a(1-t)+bt+2c=d\,\,\,\,\,\forall\,t\in\Bbb R$$

So taking succesively the values $\,t=0\,,\,t=1\,,\,t=-1\,$ , we get the linear system:

$$a-c=d$$

$$a+b+3c=d$$

$$b+2c=d$$

$$-2a-b+2c=d$$

Substracting eq. 3 from 2 we get: $\,a+c=0\Longleftrightarrow a=-c\,$ , and together with eq. 1 we get $\,2a=d\,$...and etc.

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First, we want to find 3 points in the plane. One, $(1, 0, -1)$, is given in the question. Two more can be found by plugging in arbitraty values for t: $$t = 0 \Rightarrow (1, 0, 2)$$ $$t = 1 \Rightarrow (0, 1, 2)$$

Using these 3 points, we can find 2 vectors in the plane: $$(1,0,-1) - (1,0,2) = \langle0,0,-3\rangle$$ $$(1, 0, -1) - (0, 1, 2) = \langle1, -1, -3\rangle$$

Now we take the cross product of <0, 0, -3> and <1, -1, -3> to find the normal vector to the plane, <-3, -3, 0>.

We now have an equation for a plane, $-3x -3y = c$, which is parallel to the desired plane. Finally, plug in a point on the plane to solve for c: $$-3*1 -3*0 = -3$$

Our final equation for the plane is: $$-3x-3y=-3$$ which is equal to: $$-x-y=-1$$

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