Question

If a.b.c are noncoplanar find the point of intersection of the line passing through the points $2a+3b-c,3a+4b-2c$ with the line joining the points $a-2b+3c,d-6b+bc$

Answer 1

Let $\overrightarrow{OA}=2a+3b-c$, $\overrightarrow{OB}=3a+4b-2c$, $\overrightarrow{OC}=a-2b+3c$,

$\overrightarrow{OD}=a-6b+6c$

The vector equation of the line joining the points $\overrightarrow{OA},\overrightarrow{OB}$ is $\overrightarrow{r}=\overrightarrow{OA}+t(\overrightarrow{OB}-\overrightarrow{OA}),t\in R$

$=2a+3b-c+t(3a+4b-2c-2a-3b+c)$

$=2a+3b-c+t(a+b-c)$ .........$\left(1\right)$

The vector equation of the line joining the points $\overrightarrow{OC},\overrightarrow{OD}$ is

$\overrightarrow{r}=\overrightarrow{OC}+s(\overrightarrow{OD}-\overrightarrow{OC}),s\in R$

$=a-2b+3c+s(a-6b+6c-a+2b-3c)$

$=a-2b+3c+s(-4b+3c)$ ........$\left(2\right)$

from $\left(1\right)$, we have

$\overrightarrow{r}=(2+t)\overrightarrow{a}+(3+t)\overrightarrow{b}+(-1-t)\overrightarrow{c}$

and from $\left(2\right)$, we have

$\overrightarrow{r}=(1-a)a+(-2-4s)b+(3+3s)c$

$=a+(-2-4s)b+(3+3s)+c$

comparing $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ coefficients on both sides, we get

$2+t=1\Rightarrow t=-1$

$3+t=-2-4s\Rightarrow 3-1=-2-4s$

$\Rightarrow s=-1$

Put $t=-1$ in $\left(1\right)$

$\overrightarrow{r}=2a+3b-c-1(a+b-c)$

$=2a+3b-c-a-b+c$

$=a+2b$.