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Neo92
 3 years ago
Find the equation of the plane passing through the intersection of the planes 2xy=0 and 3zy=0 and perpendicular to the plane 3x+6y3z=8.
Neo92
 3 years ago
Find the equation of the plane passing through the intersection of the planes 2xy=0 and 3zy=0 and perpendicular to the plane 3x+6y3z=8.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I hope you're prepared for some real number crunching :D It might help to get the equations of the line upon which the planes 2xy=0 and 3zy=0 intersect ^.^

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.1@Neo92 Do you have any ideas how to solve this?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let 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 soughtafter items will help us achieve... the equation of the plane :D

Neo92
 3 years ago
Best ResponseYou've already chosen the best response.0@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 2x3z?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0equation of line of intersection will be 2x = y = 3z

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Lines 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?

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.1Ok. 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Equation of plane passing through 2xy = 0 and 3zy = 0 will be: \((2xy) + \lambda(3zy) = 0\)

Neo92
 3 years ago
Best ResponseYou've already chosen the best response.0@klimenkov Yep i know. But im supposed to do this using Coordinate geometry..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here we are to just find what is \(\lambda\)..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0On second thought, @klimenkov is right, who needs lines, when you only need the vector :D Haha, my bad :3

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0There from that equation just separate x, y and z terms..

Neo92
 3 years ago
Best ResponseYou've already chosen the best response.0@waterineyes i got it. So hw cn we find

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.1What will be normal vectors for planes \(2xy=0\) and \(3zy=0\) ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(2x + y(1  \lambda) + z(3 \lambda) = 0\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here's a hint, @Neo92 The normal vector for the plane ax + by + cz = d is the vector <a , b , c>

Neo92
 3 years ago
Best ResponseYou've already chosen the best response.0@klimenkov <2,1,0>,<0,1,2> @PeterPan Thnks.

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.1Is it true that if you make a cross product of them it will be a vector in the plane that you are finding?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That'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 surefire way to get a new vector that's perpendicular to both?

Neo92
 3 years ago
Best ResponseYou've already chosen the best response.0@klimenkov Nope. Itz perpendicular to both planes. I think thtz sa answer for @PeterPan s ques also..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0As 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.. :(

Neo92
 3 years ago
Best ResponseYou've already chosen the best response.0@waterineyes You got the general eqn for line of intersection of two planes. am I correct?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, @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?

Neo92
 3 years ago
Best ResponseYou've already chosen the best response.0@waterineyes Can u xplain a little bit? Im totally new to this stuff?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But I am not quite sure that I got right answer or not...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I have applied the right formula with right calculations and I got 2xy = 0 or 2x = y.. This should be correct I believe..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'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?

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.1@PeterPan Do you see the way I want to solve this task?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0See, 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..

Neo92
 3 years ago
Best ResponseYou've already chosen the best response.0@PeterPan I got to do this using coordinate geometry pal.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here 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..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Does that mean no vectors? @Neo92

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Suppose 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: \[a_1a_2 + b_1b_2 + c_1c_2 = 0\]

Neo92
 3 years ago
Best ResponseYou've already chosen the best response.0@PeterPan Dnt knw xactly. im confused. @waterineyes got it.. :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0By using this is I have found lambda there.. One plane has equation: \[2x + y(1  \lambda) + z(3 \lambda) = 0\] And other has: \[3x+6y3z8 = 0\] Here use the above property that I have given..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@PeterPan what you got as an final answer?? If you don't wanna tell then message me..

Neo92
 3 years ago
Best ResponseYou've already chosen the best response.0wait im working on it,,, by the way can anyone tell me a gud buk o a website to refer??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes 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..

Neo92
 3 years ago
Best ResponseYou've already chosen the best response.0@waterineyes ,@PeterPan ,@klimenkov thnx 4 da help guys..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You are welcome dear..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Might as well... You could get the cross product of the normal vectors of 2xy = 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0By 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

Neo92
 3 years ago
Best ResponseYou've already chosen the best response.0does the vector <p, q, r> means the direction cosines of the line//

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0NO, the vector <p , q , r> means the vector that is perpendicular to the plane. No lines are involved, thank heavens :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here, 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Such that < p q r > is perpendicular to the plane.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Then if we take any point (x , y , z) in the planedw:1364728184367:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1364728197885:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And then draw the vector from (a b c) to (x y z)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1364728225211:dw This vector would have be <xa , yb , zc>

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thus the dot (scalar) product of <p q r> and <xa , yb , zc> should ALWAYS be zero. \[\Large <p \ , \ q \ , \ r> \cdot <xa \ , \ yb \ , \ z c>=0\]\[\huge p(xa)+q(yb)+r(zc)=0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And that's how the equation of the plane came about ^.^

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I hope you liked my story :D

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.1Very good @PeterPan. I would have given you the second medal if I could.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's so awesome of you guys, this was so much fun ^.^
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