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

Question

This is some UNGRADED homework that I cannot solve (the answers are available to us through the Math dept. website, but I don't understand at all how to get this answer).

The question is:

Write an equation for the plane which contains the x-axis and the point (0,1,2).

Apparently, the answer is -2y+z = 0 but I don't even know where to begin. Somebody asked the professor how to solve it, but he wasn't very helpful. He's one of the genius-types that isn't the best at teaching.

In advance, THANK YOU so much!

vahucel - This is the most helpful answer I've ever gotten on Y!A. Thank you so much!

Answer 1

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.

Answer 2

Once you find the normal vector, it should look like 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.

Answer 3

i have been browsing the web more than three hours today seeking the answers to the same question, yet I haven't found a more interesting discussion like this. it's pretty worth enough for me.

Answer 4

Sounds weird

Answer 5

confusing subject. look over google or bing. it can help!