Qualification > Math
MAth DOubts
astarmathsandphysics:
is ok.
Twinkle Charms:
Q4)Two planes have equation x+2y-2z=2 and 2x-3y+6z=3. The planes intersect in the straight line L. Find a vector equtaion for line L.
ok so i just know to get the vector equation of a line we use this>> r=a+$p, where a is the position vector, $ is a scaler and p is direction vector, after that how to get all that i dunno....please help...=\
CHEMMASTER6000:
so to find the direction vector you cross multiply the normals using matrices rules. and to find the point on the line you do simultaneous equation using the two plane quations but thats where it gets tricky. the trick is to sub in a value for x y or z which will give zero thus only one unknown is left. i rather not type the answer here cause its difficult to type and i dont know how to scan in the work sheet into forum ???hope it my methods helps
Twinkle Charms:
--- Quote from: inverse tangent graph on May 23, 2010, 07:25:10 pm ---so to find the direction vector you cross multiply the normals using matrices rules. and to find the point on the line you do simultaneous equation using the two plane quations but thats where it gets tricky. the trick is to sub in a value for x y or z which will give zero thus only one unknown is left. i rather not type the answer here cause its difficult to type and i dont know how to scan in the work sheet into forum ???hope it my methods helps
--- End quote ---
u scan ur worksheet and attach to your post cz i aint gettin what u said to do for the point on the line =\ to attach click on the additional options...
CHEMMASTER6000:
hahaa ok i think i failed at the scanning thing but il try it here .
(a)x+2y-2z=2 and (b)2x-3y+6z=3
so point on line would be . so simultaneously solve both of them . (trick) sube 0 into one of them . i choose z
when z=0
(a)x+2y=2 (b) 2x-3y=3
now its easy so you can solve for x and y. sub back into eqn and walla you get the point ahh haha sound so nerdy
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