The point if intersection of the lines $l_1: \mathbf{r}(t)=(\mathbf{i}-6 \mathbf{j}+2…
The point if intersection of the lines $l_1: \mathbf{r}(t)=(\mathbf{i}-6 \mathbf{j}+2 \mathbf{k})+t(\mathbf{i}+2 \mathbf{j}+\mathbf{k})$ $l_2: \mathbf{R}(u)=(4 \mathbf{j}+\mathbf{k})+u(2 \mathbf{i}+\mathbf{j}+2 \mathbf{k})$ is
$(4,4,5)$
$(6,4,7)$
$(8,8,9)$
$(10,12,11)$
Solution
Put $\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}$
$\therefore x \mathbf{i}+y \mathbf{j}+z \mathbf{k}=(\mathbf{i}-6 \mathbf{j}+2 \mathbf{k})+t(\mathbf{i}+2 \mathbf{j}+\mathbf{k})$
$\therefore$ Any point on line is $P(1+t,-6+2 t, 2+t)$ is satisfied the second equation of line.
$\begin{aligned} \therefore \quad & (1+t) \mathbf{i}+(-6+2 t) \mathbf{j}+(2+t) \mathbf{k} \\ & =2 u \mathbf{i}+(4+u) \mathbf{j}+(1+2 u) \mathbf{k}\end{aligned}$
On equating the coefficients of $\mathbf{i}, \mathbf{j}$ and $\mathbf{k}$, we get
$1+t=2 u$
$\begin{aligned} & \Rightarrow \quad t-2 u+1=0 \\ & -6+2 t=4+u \\ & \Rightarrow \quad 2 t-u-10=0 \\ & \text { and } 2+t=1+2 u \\ & \Rightarrow \quad t-2 u+1=0 \\ & \end{aligned}$
On solving Eqs. (i) and (ii), we get
$\begin{gathered}t=7, \quad u=4 \\ \therefore \quad P(1+7,-6+2 \times 7,2+7)=P(8,8,9)\end{gathered}$