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equation of line of intersection will be 2x = y = 3z

Equation of plane passing through 2x-y = 0 and 3z-y = 0 will be:
\((2x-y) + \lambda(3z-y) = 0\)

@klimenkov Yep i know. But im supposed to do this using Coordinate geometry..

Here we are to just find what is \(\lambda\)..

There from that equation just separate x, y and z terms..

@waterineyes i got it. So hw cn we find

What will be normal vectors for planes \(2x-y=0\) and \(3z-y=0\) ?

\(2x + y(-1 - \lambda) + z(3 \lambda) = 0\)

@waterineyes You got the general eqn for line of intersection of two planes. am I correct?

@waterineyes Can u xplain a little bit? Im totally new to this stuff?

But I am not quite sure that I got right answer or not...

@waterineyes Yep..

@PeterPan Dnt knw xactly. im confused.
@waterineyes got it.. :D

wait im working on it,,, by the way can anyone tell me a gud buk o a website to refer??

2x-y=0 ??

Great...

You are welcome dear..

You are welcome.

which is normal to your plane, the equation of the plane is given by p(x - a) + q(y - b) + r(z - c) = 0

does the vector

means the direction cosines of the line//

means the vector that is perpendicular to the plane. No lines are involved, thank heavens :D

|dw:1364728078914:dw|

Such that < p q r > is perpendicular to the plane.

Then if we take any point (x , y , z) in the plane|dw:1364728184367:dw|

|dw:1364728197885:dw|

And then draw the vector from (a b c) to (x y z)

|dw:1364728225211:dw|
This vector would have be

as you can see.

and

\cdot

And that's how the equation of the plane came about ^.^

I hope you liked my story :D

That's so awesome of you guys, this was so much fun ^.^

Thz z cool guys....