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Find a unit vector that is normal to the line with equation 2x-5y=3.

Feb 17, 2020

$$2x-5y=3\\ y = \dfrac{2x-3}{5} = \dfrac 2 5 x - \dfrac 3 5\\ m = \dfrac 2 5\\ m^\perp= -\dfrac{1}{m} = -\dfrac 5 2\\ n = \left(1, -\dfrac 5 2\right)\\ \hat{n} = \dfrac{2}{\sqrt{29}}\left(1,-\dfrac 5 2\right) = \left(\dfrac{2}{\sqrt{29}}, -\dfrac{5}{\sqrt{29}}\right)$$