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 $a-2b+3c,a-6b+6c.$

Answer 1

$\overrightarrow{r}=\overrightarrow{a}+\lambda (\overrightarrow{b}-\overrightarrow{a})$ (equation of a line)

So, the 2 intersecting lines are,

$\overrightarrow{r_1}=(2a+3b-c)+\lambda \left(\right(3a+4b-2c)-(2a+3b-c\left)\right)=(2a+3b-c)+\lambda (a+b-c)$

and $\overrightarrow{r_2}=(a-2b+3c)+\mu \left(\right(a-6b+6c)-(a-2b+3c\left)\right)=(a-2b+3c)+\mu (-4b+3c)$

At the point of intersection, $\overrightarrow{r_1}=\overrightarrow{r_2}$.

ie, $(2a+3b-c)+\lambda (a+b-c)=(a-2b+3c)+\mu (-4b+3c)$

As $a,b,c$ are non-coplanar, there could only be one linear combination of the vectors for each spanning vector. (characteristic of vector space)

ie, the coefficients of $a,b,c$ are unique.

From equation of $\overrightarrow{r_2}$ we can see that whatever the value of $\mu$, the coefficient of $a$ is constant and is $1$.

But for $\overrightarrow{r_1}$ only $\lambda =(-1)$ gives coefficient $1$ for $a$.

So, the intersection point $\overrightarrow{r}=(2a+3b-c)+(-1)(a+b-c)=(a+2b)$