Question

The intersection point of lines ...
$\overrightarrow{r}=i+2j+3k+\lambda (2i+3j+4k)$
$\overrightarrow{r}=4i+j+\mu (5i+2j+k)$ is

Answer 1

The intersection point of lines

$r=i+2j+3k+\lambda (2i+3j+4k)$

$r=4i+j+\mu (5i+2j+k)$

Solution: Equating terms

$(1+2\lambda )i+(2+3\lambda )j+(3+4\lambda )k=r$

$(4+5\mu )ii+(1+2\mu )j+(0+\mu )k=r$

$1+2\lambda =4+5\mu$

$2+3\lambda =1+2\mu$

$3+4\lambda =\mu$ ………$\left(1\right)$

From $\left(1\right)$

$1+2\lambda =4+5(3+4\lambda )$

$\Rightarrow 1+2\lambda =4+15+20\lambda$

$-18\lambda =18$

$\lambda =-1$

From $\left(1\right)$

$3+4(-1)=\mu$

$\mu =-1$

Substituting in r

$r=i+2j+3k+(-2i)+(-3j)+(-4k)$

$=-i-j-k$

$r=4i+j+(-5i)+(-2j)+(-k)$

$=-i-j-k$

$\therefore$ Point of intersection $(-1,-1,-1)$.