Question

3x-3=3y-6 =z-7,find the unit vector parallel to the above line

Answer 1

Let us rewrite the equation in form

$\dfrac{x-1}{\frac{1}{3}} = \dfrac{y-2}{\frac{1}{3}} = \dfrac{z-7}{1}$ .

In such form the denominators represent the coordinates of the vector parallel to the line. The coordinates are (1/3, 1/3, 1). But we can see the vector is not unit, so we should normalize it.

The length of the vector above is

$\sqrt{(1/3)^2+(1/3)^2+1^2} = \dfrac{\sqrt{11}}{3}.$

Therefore, the unit vector will have coordinates $\left(\dfrac{1/3}{\sqrt{11}/3}\,,\dfrac{1/3}{\sqrt{11}/3}\,, \dfrac{1}{\sqrt{11}/3}\right)$ , after simplifying

$\left( \dfrac{\sqrt{11}}{11}\,, \dfrac{\sqrt{11}}{11}\,, \dfrac{3\sqrt{11}}{11}\right)$ .