Cal 3: Find equation of the plane containing (0,1,2) and the x-axis?

1) A plane has an equation ax+by+cz+d=0, in which (a,b,c) is a normal vector(perpendicular to the plane) and d is the distance from the plane to the origin (0,0,0)

2) As your plane contains x-axis, then it contains the point (0,0,0) and you have a.0 + b.0 + c.0 + d = 0... then d =0 and the equation changes to ax+by+cz=0

3) As your plane contains x-axis, then the normal vector is always perpendicular to x-axis... that means the perpendicular vector is of the form (0,b,c)... then the equation changes again to by + cz = 0

4) As the plane contains the point (0,1,2), then this point must verify the equation... we have b.1 + c.2 = 0

Now you may choose the numbers b and c that verify the last equation... b + 2c = 0... choose the easiest numbers... for example b=2 and c= -1

5) Replacing this numbers in equation obtained in 3) we get

2y -z =0... the same if you change the signals -2y +z =0.

Once you find the normal vector, it should look like <a,b,c> or some other similar notation. Then the plane's equation is ax + by + cz + d = 0, where the a,b, and c are just the components of the normal vector. Then plug the known point into x,y, and z in the equation to determine the value of d. If you want, you can solve it for z, but usually the equation of planes are left in the above format so one could easily derive the normal vector the other way just by looking at it.

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Sounds weird

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