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.!