Let $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be three non-coplanar vectors. The vector equation of a line…
Let $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be three non-coplanar vectors. The vector equation of a line which passes through the point of intersection of two lines, one joining the points $\mathbf{a}+2 \mathbf{b}-5 \mathbf{c}$, $-\mathbf{a}-2 \mathbf{b}-3 \mathbf{c}$ and the other joining the points $-4 \mathbf{c}, 6 \mathbf{a}-4 \mathbf{b}+4 \mathbf{c}$ is
Equation of line joining the points,
$\mathbf{a}+2 \mathbf{b}-5 \mathbf{c},-\mathbf{a}-2 \mathbf{b}-3 \mathbf{c}$ is
$
\mathbf{r}=(\mathbf{a}+2 \mathbf{b}-5 \mathbf{c})+\lambda_1,(2 \mathbf{a}+4 \mathbf{b}-2 \mathbf{c})
$
Similarly, equation of the line joining the points $-4 c, 6 a-4 b+4 c$ is
$
\mathbf{r}=-4 \mathbf{c}+\lambda_2(6 \mathbf{a}-4 \mathbf{b}+8 \mathbf{c})
$
Now, for point of intersection of lines (i) and (ii)
$
\begin{aligned}
\left(2 \lambda_1+1\right) \mathbf{a}+ & \left(4 \lambda_1+2\right) \mathbf{b}+\left(-2 \lambda_{,}-5\right) \mathbf{c} \\
= & \left(6 \lambda_2\right) \mathbf{a}+\left(-4 \lambda_2\right) \mathbf{b}+\left(8 \lambda_2+4\right) \mathbf{c}
\end{aligned}
$
On comparing
$
\begin{aligned}
2 \lambda_1+1 & =6 \lambda_2 \\
4 \lambda_1+2 & =-4 \lambda_2 \\
-2 \lambda_1-5 & =8 \lambda_2+4
\end{aligned}
$
and
From Eqs. (iii), (iv) and (v)
$
\lambda_1=-\frac{1}{2} \text { and } \lambda_2=0
$
So, intersection points is $-4 \mathbf{c}$ and for $\mu=-3$ the line $\mathbf{r}=(3 \mathbf{a}+6 \mathbf{b}-\mathbf{c})+\mu(\mathbf{a}+2 \mathbf{b}+\mathbf{c})$ passes through point -4 conly