Answer to Question #324968 in Analytic Geometry for Agenda

Question

Given that A(2,3,-1) and B(1,2,4) are points in space. Find the unit vector along the direction of AB and the one along BA. Also find the position vector of the point C which is on the mid-point of AB

Answer 1

The vector $\overrightarrow{AB}$ has the form: $\overrightarrow{AB}=(-1,-1,5)$. The length of the vector is: $|\overrightarrow{AB}|=\sqrt{27}$. Thus, the unit vector along the direction of $AB$ has the form: $(-\frac{1}{\sqrt{27}},-\frac{1}{\sqrt{27}},\frac{5}{\sqrt{27}})$. The opposite unit vector has the form: $(\frac{1}{\sqrt{27}},\frac{1}{\sqrt{27}},-\frac{5}{\sqrt{27}})$. Find coordinates of the point $C:$ $C=(2-\frac12,3-\frac12,-1+\frac52)=(1.5,2.5,1.5)$. It is not mentioned, what is the origin of the vector. Consider two options: 1. $O=(0,0,0)$ is the origin, Then, $\overrightarrow{OC}=(1.5,2.5,1.5)$. It is the position vector. 2. Suppose that $A=(2,3,-1)$ is the origin. Then, $\overrightarrow{AC}=(-0.5,-0.5,2.5)$.