Here's the question you clicked on:
Neo92
Find the equation of the plane passing through the intersection of the planes 2x-y=0 and 3z-y=0 and perpendicular to the plane 3x+6y-3z=8.
I hope you're prepared for some real number crunching :D It might help to get the equations of the line upon which the planes 2x-y=0 and 3z-y=0 intersect ^.^
@Neo92 Do you have any ideas how to solve this?
Let nothing distract you from the ultimate prize ^.^ We're after a point on the plane, as well as a normal vector, a vector which is perpendicular to the plane. These highly sought-after items will help us achieve... the equation of the plane :D
@klimenkov No pal.. Can u tell me a gud site or a buk to refer? @PeterPan What is the eqn of the line of intersection? is it 2x-3z?
equation of line of intersection will be 2x = y = 3z
Lines are fickly creatures :) While in 2d, they are as docile and manageable as anything, in 3d, they are tricksters, ready to confuse the unsuspecting student. For in 3d, one equation is not sufficient to define a line... you need 2. A line is always defined as the intersection of two planes, there's no avoiding it! Now, all you need to get the equationS of a line is a point on that line, and a vector direction :) Any idea how to get those?
Ok. I will help you to solve, if I see that you are interested in. Actually there is no need in the equation of the line of intersection. Do you know what is a vector product?
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 \(\lambda\)..
On second thought, @klimenkov is right, who needs lines, when you only need the vector :D Haha, my bad :3
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 \(3z-y=0\) ?
\(2x + y(-1 - \lambda) + z(3 \lambda) = 0\)
Here's a hint, @Neo92 The normal vector for the plane ax + by + cz = d is the vector <a , b , c>
@klimenkov <2,-1,0>,<0,-1,2> @PeterPan Thnks.
Is it true that if you make a cross product of them it will be a vector in the plane that you are finding?
That's right :) What we know about the plane we're looking for, it's gonna contain the entire line of intersection of the two planes you just found normal vectors for. That line is going to be orthogongal (perpendicular) to both their normal vectors, then. Given two vectors, what's a sure-fire way to get a new vector that's perpendicular to both?
@klimenkov Nope. Itz perpendicular to both planes. I think thtz sa answer for @PeterPan s ques also..
As this is perpendicular to the plane 3x + 6y - 3z - 8 = 0 So: \(6 -6 - 6\lambda -9\lambda = 0\) \(-15 \lambda = 0\) \(\lambda = 0..\) SO: Equation required should be : 2x - y = 0 Or I am wrong.. :(
@waterineyes You got the general eqn for line of intersection of two planes. am I correct?
No, @Neo92 Your two vectors <2 , -1 , 0> and <0 , -1 , 2> are orthogonal (perpendicular) to your planes, but these vectors are definitely not in the plane we're looking for. What WILL be in the plane we're looking for, is the vector that's orthogonal (perpendicular) to both <2 , -1 , 0> and <0, -1 , 2> What's that vector?
@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...
I have applied the right formula with right calculations and I got 2x-y = 0 or 2x = y.. This should be correct I believe..
I'm just saying, guys @waterineyes and @Neo92 I found out the hard way (through @klimenkov 's friendly reminder) that the line of intersection is actually not necessary, only the vector. Maybe this is taking it a bit too far?
@PeterPan Do you see the way I want to solve this task?
See, the equation of plane passing through the intersection of tow other planes namely \(a_1x + b_1y + c_1z + d_1 = 0\) and \(a_2x + b_2y + c_2z + d_2 = 0\) is given by: \[(a_1x + b_1y + c_1z + d_1) + \lambda(a_2x + b_2y + c_2z + d_2) = 0\] Where \(\lambda \) is some constant..
@PeterPan I got to do this using coordinate geometry pal.
Here we have to find lambda and for that it is necessary that you must be provided with one more constraint which in this case is our required plane is perpendicular to another plane, this is what helps in finding value of lambda..
Does that mean no vectors? @Neo92
Suppose the two planes are : \[a_1x + b_1y + c_1z + d_1 = 0 \quad and \quad a_2x + b_2y + c_2z + d_2 = 0\] So if two planes are perpendicular then: \[a_1a_2 + b_1b_2 + c_1c_2 = 0\]
@PeterPan Dnt knw xactly. im confused. @waterineyes got it.. :D
By using this is I have found lambda there.. One plane has equation: \[2x + y(-1 - \lambda) + z(3 \lambda) = 0\] And other has: \[3x+6y-3z-8 = 0\] Here use the above property that I have given..
@PeterPan what you got as an final answer?? If you don't wanna tell then message me..
wait im working on it,,, by the way can anyone tell me a gud buk o a website to refer??
Yes this is correct.. So remember when a plane is passing through intersection of two other planes and is perpendicular to some other plane, then you will find equation like this..
@waterineyes ,@PeterPan ,@klimenkov thnx 4 da help guys..
You are welcome dear..
Might as well... You could get the cross product of the normal vectors of 2x-y = 0 and 3z - y = 0 <2 , -1 , 0> x <0 , -1, 3> = -<3 , 6 , 2> and this is one of the vectors on your plane the other vector is the normal vector of 3x + 6y - 3z = 8 which is <3 , 6 , -3> Now, since both <3 , 6, 2> and <3 , 6 , -3> are on the plane, their cross product, which is <-30 , 15 , 0> is NORMAL to the plane you're looking for. Now you just have to find a point on your plane, which can be done by letting x = 0, and you'll find that y = 0 and z = 0, using the equations 2x - y = 0 and 3z - y = 0 Thus, the point (0 , 0 , 0) is on your plane. Now that you have your point and normal vector, the equation of the plane is -30(x - 0) + 15(y - 0) + 0(z - 0) = 0 -30x + 15y = 0 -2x + y = 0
By the way, given a point (a , b , c) in your plane and a vector <p , q , r> 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 <p, q, r> means the direction cosines of the line//
NO, the vector <p , q , r> means the vector that is perpendicular to the plane. No lines are involved, thank heavens :D
Here, I'll derive it for you :) Say you have a plane, we don't know its equations yet. And a point (a , b , c) on the plane, as well as a vector <p , q , r> |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|
And then draw the vector from (a b c) to (x y z)
|dw:1364728225211:dw| This vector would have be <x-a , y-b , z-c>
No matter where the point (x y z) is, as long as it's on the plane, the vector from (a b c) to (x y z) would ALWAYS be perpendicular to the vector <p q r> as you can see.
Thus the dot (scalar) product of <p q r> and <x-a , y-b , z-c> should ALWAYS be zero. \[\Large <p \ , \ q \ , \ r> \cdot <x-a \ , \ y-b \ , \ z -c>=0\]\[\huge p(x-a)+q(y-b)+r(z-c)=0\]
And that's how the equation of the plane came about ^.^
I hope you liked my story :D
Very good @PeterPan. I would have given you the second medal if I could.
That's so awesome of you guys, this was so much fun ^.^