My question is with respect to PS3 part 2 #3a. I got to the point where x=at, y=bt, z=(-a-bt). They claim there is a redundancy in the line. I don't see why. x, y, and z seem to be three distinct, valid equations. I'll attach the problem and their solution in my next post. Thanks in advance.

Question

My question is with respect to PS3 part 2 #3a. I got to the point where x=at, y=bt, z=(-a-bt). They claim there is a redundancy in the line. I don't see why. x, y, and z seem to be three distinct, valid  equations. I'll attach the problem and their solution in my next post. Thanks in advance.

anonymous · Answer

Attached is the problem and their solution.

baru · Answer

x,y,z are not three distinct equations,
the are the equations for the three components of the position vector...
 
what i mean is
let L be any line on the given plane.
let P be any point on L

then OP= <at , bt  , -a-bt >

baru · Answer

the redundancy part is a bit harder to explain... i'll try to explain it this way if the direction vector of a line on the plane has components then it has to satisfy the relation a+2b+c=0 look at the equation a+2b+c=0 if you choose any value for two of the components, the value of the third is automatically determined. so if i choose values for a and b, then c automatically becomes -a-2b lets say i choose these values for a and b a= $\alpha$ b=1 then the equation becoems $\alpha$+2(1)+c=0 so, c=-$\alpha$-2 so your direction vector becomes <$\alpha$ , 1 , -$\alpha$-2> notice how we have reduced this to a single parameter. you have to choose only $\alpha$, and the direction vector is decided. the required equation is simply any scalar multiple of this vector, thus we get <$\alpha$ , 1 , -$\alpha$-2> $\times t$

anonymous · Answer

Thanks again.