Let the line be y=3x+4
Let's find two unit vectors that are parallel to that line.
The slope the given line is 3.
Suppose that, this line makes an angle of θ \theta θ with the +ve direction of the X−Axis.
Then, the unit vector u ‾ = ( cos θ , sin θ ) \overline{u}=(\cos\theta,\sin\theta) u = ( cos θ , sin θ ) is parallel to the line.
By the definition of slope, we have
tan θ = 3 \tan\theta=3 tan θ = 3
therefore
sec 2 θ = 1 + tan 2 θ = 10 \sec^2\theta=1+\tan^2\theta=10 sec 2 θ = 1 + tan 2 θ = 10
c o s θ = 1 sec θ = ± 1 10 cos\theta=\frac{1}{\sec\theta}=\pm\frac{1}{\sqrt{10}} cos θ = s e c θ 1 = ± 10 1
sin θ = 1 − c o s 2 θ = 1 − 1 10 = ± 3 10 \sin\theta=\sqrt{1-cos^2\theta}=\sqrt{1-\frac{1}{10}}=\pm\frac{3}{\sqrt{10}} sin θ = 1 − co s 2 θ = 1 − 10 1 = ± 10 3
Hence u ‾ ( 1 10 , 3 10 ) \overline{u}(\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}) u ( 10 1 , 10 3 ) and v ‾ ( − 1 10 , − 3 10 ) \overline{v}(-\frac{1}{\sqrt{10}},-\frac{3}{\sqrt{10}}) v ( − 10 1 , − 10 3 ) are two vectors, parallel to the line y=3x+4
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