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 $(\pm \frac{1}{\sqrt{10}}, \pm \frac{3}{\sqrt{10}}) .$ $(+-1/sqrt10,+-3/sqrt10).$

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

Suppose that, this line makes an angle of $θ$ $theta$ with the $+ v e$ $+ve$

direction of the $X -$ $X-$ Axis, where, $θ \in (0, π) - {\frac{π}{2}} .$ $theta in (0,pi)-{pi/2}.$

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

by, $\to u = (cos θ, sin θ) .$ $vecu=(costheta, sintheta).$

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

$tan θ = 3, θ \in (0, π) - {\frac{π}{2}} .$ $tan theta=3, theta in (0,pi)-{pi/2}.$

$But, tan θ > 0 \Rightarrow 0 < θ < \frac{π}{2} .$ $"But, "tan theta gt 0 rArr 0 lt theta lt pi/2.$

${sec}^{2} θ = 1 + {tan}^{2} θ = 1 + 9 = 10 \Rightarrow sec θ = \pm \sqrt{10} .$ $sec^2theta=1+tan^2theta=1+9=10 rArr sectheta=+-sqrt10.$

$θ \in (0, \frac{π}{2}) \Rightarrow cos θ = \frac{1}{sec θ} = + \frac{1}{\sqrt{10}} .$ $theta in (0,pi/2) rArr costheta=1/sectheta=+1/sqrt10.$

Also, $sin θ = tan θ sec θ = + \frac{3}{\sqrt{10}} .$ $sintheta=tanthetasectheta=+3/sqrt10.$

Hence, $\to u = (\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}}) .$ $vec u=(1/sqrt10, 3/sqrt10).$

The other vector parallel to the line is $- \to u .$ $-vecu.$

Enjoy Maths.!