for vector $\left<a,b\right>$, the unit vector is $\left< \dfrac{a}{\sqrt{a^2+b^2}} ,\dfrac{b}{\sqrt{a^2+b^2}} \right>$
oh, and refrain from including links to pay sites like "studydaddy", please.
for vector $\left<a,b\right>$, the unit vector is $\left< \dfrac{a}{\sqrt{a^2+b^2}} ,\dfrac{b}{\sqrt{a^2+b^2}} \right>$
oh, and refrain from including links to pay sites like "studydaddy", please.
I thought I had already responded to this.
First, "the unit vector to <a,b>" means the unit vector in the same direction as <a,b>.
Taking $\theta$ to be the angle the vector <a, b> makes with the positive x-axis, $tan(\theta)= \frac{b}{a}$
But also $\frac{\frac{b}{\sqrt{a^2+b^2}}}{\frac{a}{\sqrt{a^ 2+b^2}}}= \frac{b}{a}$
In order that this be a unit vector, its length must be 1.
$\sqrt{\left(\frac{a}{\sqrt{a^2+b^2}}\right)^2+ \left(\frac{b}{\sqrt{a^2+b^2}}\right)^2}$$= \sqrt{\frac{a^2}{a^2+b^2}+ \frac{b^2}{a^2+b^2}}= \sqrt{\frac{a^2+b^2}{a^2+b^2}}= 1$