Question

Show that the points $O(0,0),A(2,-3,3),B(-2,3,-3)$ are collinear. Find the ratio in which each point divides the segment joining the other two.

Answer 1

Let's find area of $△OAB$ which is given by,

$\Delta =\frac{1}{2}\left|OA\right|\left|OB\right|\sin \theta$

where $\theta$ is angle between $\overrightarrow{OA}$ and $\overrightarrow{OB}$

Now, $\cos \theta =\frac{\overrightarrow{OA}.\overrightarrow{OA}}{\left|\overrightarrow{OA}\right|\left|\overrightarrow{OB}\right|}$

$=\frac{-4-9-9}{\sqrt{4+9+9}\sqrt{4+9+9}}=\frac{-22}{22}$

$\Rightarrow$ $\cos \theta =-1\Rightarrow$ $\theta ={180}^o$

$\Rightarrow$ $\Delta =0$ as $\sin 180=0$

$\Rightarrow$ $O,A,B$ are collinear

Now, let $A$ divides $\overrightarrow{OB}$ in $k:1$

$\Rightarrow$ $2=\frac{-2k+0}{k+1}\Rightarrow$ $2k+2=-2k\Rightarrow$ $4k=-2\Rightarrow$ $k=-\frac{1}{2}$

$\Rightarrow$ $A$ divides $\overrightarrow{OB}$ in $1:2$ externally

Now for $B$

$\Rightarrow$ $-2=\frac{2k+0}{k+1}\Rightarrow$ $-2k-2=2k\Rightarrow$ $4k=-2$

$\Rightarrow$ $k=-\frac{1}{2}$

$\Rightarrow$ $B$ divides $\overrightarrow{OA}$ in $1:2$ externally for $O$

$\Rightarrow$ $O=\frac{-2k+2}{k+1}\Rightarrow$ $2k=2\Rightarrow$ $k=1$

$\Rightarrow$ $O$ divides $\overrightarrow{AB}$ in $1:1$ internally