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March 28, 2015

March 28, 2015

Posted by **Anita** on Monday, April 9, 2012 at 3:56am.

- Calculus and vectors -
**Reiny**, Monday, April 9, 2012 at 9:10amthe normal of the given plane is

(3,0,-5) which is then the direction of your line

A vector equation of that line is

r = (x,y,z) = (2,9,-3) + t(3,0,-5) where t is your parameter

x = 2 + 3t ---> t = (x-2)/3

y = 9 + 0t ---> y = 9

z = -3 - 5t --> t = (z+3)/-5

so

(x-2)/3 = (z+3)/-5 , y = 9

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