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Math Help

both appear to be wrong.

what is the goal of 1?

and in 2, you find the distance the same way you do it in 2d. use the distance formula, then minimize it (differentiate, solve for 0s which gives you a min). just expand it into 3d.
 
For number 2 whats wrong with useing the magnitude of the projection of x on to a to find the shortest distance. The goal of number one is to solve the DE.

The name of the first class is Differential Equations, the name of the second class is Linear Algebra.

 
Number 2 I just did it a different way and I got the same answer. Like I pointed out that it was really 2x1-x2+3x3=6 not 2x-y+3z=6. I change it because I thought it would be easier for you people to read.

Now the other way was like this vector x=(0,0,0)+t(vector a), vector x=(2t,-t,3t) x1=2t x2=-1, and x3=3t I plug this back into the original problem.
4t+t+9t=6 solve for t you get t=3/7. Plug this back into x1, x2, x3 and you get (6/7,-3/7,9/7) The same answer I got before.
 
For number 2 I know I am right. I asked me math teacher today about it and he said the way I did it was fine. But number 1 I still don't know if I am doing it right.
 
Jet the question was to solve the DE.


the DE was dy/dx=sin(x+y)

I got K=tan(x+y)-sec(x+y)-x I want to know if that is correct.
 
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