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\thetaθ with the +ve direction of the X−Axis.
Then, the unit vector u‾=(cosθ,sinθ)\overline{u}=(\cos\theta,\sin\theta)u=(cosθ,sinθ) is parallel to the line.
By the definition of slope, we have
tanθ=3\tan\theta=3tanθ=3
therefore
sec2θ=1+tan2θ=10\sec^2\theta=1+\tan^2\theta=10sec2θ=1+tan2θ=10
cosθ=1secθ=±110cos\theta=\frac{1}{\sec\theta}=\pm\frac{1}{\sqrt{10}}cosθ=secθ1=±101
sinθ=1−cos2θ=1−110=±310\sin\theta=\sqrt{1-cos^2\theta}=\sqrt{1-\frac{1}{10}}=\pm\frac{3}{\sqrt{10}}sinθ=1−cos2θ=1−101=±103
Hence u‾(110,310)\overline{u}(\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}})u(101,103) and v‾(−110,−310)\overline{v}(-\frac{1}{\sqrt{10}},-\frac{3}{\sqrt{10}})v(−101,−103) are two vectors, parallel to the line y=3x+4
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments