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Zenmo
 one year ago
@Small Question, don't need to solve@
Find the general form of the equation of the plane with the given characteristics.
Passes through (2, 2, 1) and (1, 1, 1) and is perpendicular to 2x3y+z=3.
Zenmo
 one year ago
@Small Question, don't need to solve@ Find the general form of the equation of the plane with the given characteristics. Passes through (2, 2, 1) and (1, 1, 1) and is perpendicular to 2x3y+z=3.

This Question is Closed

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.1For the points (2, 2, 1) and (1, 1, 1). How do I know, which point is the initial point? Whichever z variable value is greater?

dan815
 one year ago
Best ResponseYou've already chosen the best response.1so firsst find your normal line

dan815
 one year ago
Best ResponseYou've already chosen the best response.1mmaybe if i show u visually it will be easy for u to do it yourself

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.1So if the solution in the book is 7x+y11z5=0, but I put the answer as 7xy+11z+5=0. I would still be correct?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3x = 2 is equivalent to x = 2

dan815
 one year ago
Best ResponseYou've already chosen the best response.1you want to find your normal line of the plane, move that normal like to a point, then see the other line connecting p1 and p2, then u have 2 lines and u can define your plane now

dan815
 one year ago
Best ResponseYou've already chosen the best response.1u can take the cross of these 2 lines to get your new normal <a,b,c> then the plane equation is ax+by+cz=d and u can find d but subbing in one of your points <x,y,z?

dan815
 one year ago
Best ResponseYou've already chosen the best response.1if that plane quations seems a little magical, u can get it more lgoically by computing the dot product of your normal line with one of your vectors in your plane, <xx1,yy1,zz1> <a,b,c> dot <xx1,yy1,zz1> =0 will also yield the same plane equation

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[u\times v\] gives the normal vector right

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.1ok, I got it now. Thanks for clarification. X = 2 is the same as X = 2. Which, is the same as u x v =  ( v x u).

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Yes, direction of normal vector is not affected by multiplying the vector by a nonzero scalar

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.1But, just to make sure. I can put the answer as 7xy+11z+5=0, even though the solution is 7x+y11z5=0?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3both are same, but `7x+y11z5=0` looks more neat

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.314x+2y2z10=0 also works

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.1so to get 7x+y11z5=0. I would have to start with the point (2, 2, 1) then. Which, comes back to the question of what initial point to use.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3it is an equation, multiplying the same thing both sides doesn't change it in mathematically

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.1Or, whenever I get a negative constant for variable X, I'll just divide it by 1?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3why do you care what point you start wid

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3you end up wid same answer either way right

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.1Yea, but the signs would be different, hence: 7xy+11z+5=0 vs 7x+y11z5=0. So, either format works, but the positive one looks neater.

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.1Just really really making sure

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3okay i got you, then just multiply 1 in the end to make it look neat :)
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