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Consider the line that passes through the point and is parallel to the given vector. (1, -5, 9) ‹-1, 2, -3› (a) Find symmetric equations for the line.
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x= -1t+1 y=2t-5 z= -3t+9
this is parametric, we want symmetric,
x-1=-t -x+1=t
y+5=2t (y+5)/2 =t
so, how do you get parametric eq's out of that? and then symmetric is just parametric solved for t?
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yes, solve for t to get symmetric
Example ------------ vector <a,b,c> point <p,q,r> x= at +p y= bt+q z= ct+ r
awesome.
so t=(z-9)/-3
yes
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yep, a line is just a vector add to a point.|dw:1314982153106:dw|
since the vector can be any scaled version of itself; we use a "t" to indicate that any size of this vector is sufficient to produce a line
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