How do you find the two unit vectors in R^2 parallel to the line y=3x+4?

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Answer 1 · 2017-02-24T14:16:50

Reqd. vectors are #(+-1/sqrt10,+-3/sqrt10).#

Note that the slope the given line is #3.# So, it is not vertical.

Suppose that, this line makes an angle of #theta# with the #+ve#

direction of the #X-#Axis, where, #theta in (0,pi)-{pi/2}.#

Clearly, then, the Unit vector #vec u# parallel to the line is given

by, #vecu=(costheta, sintheta).#

Now, by the Defn. of Slope, we have,

#tan theta=3, theta in (0,pi)-{pi/2}.#

#"But, "tan theta gt 0 rArr 0 lt theta lt pi/2.#

#sec^2theta=1+tan^2theta=1+9=10 rArr sectheta=+-sqrt10.#

#theta in (0,pi/2) rArr costheta=1/sectheta=+1/sqrt10.#

Also, #sintheta=tanthetasectheta=+3/sqrt10.#

Hence, #vec u=(1/sqrt10, 3/sqrt10).#

The other vector parallel to the line is #-vecu.#

Enjoy Maths.!