Question

If the lines
$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then find the value of $k$.

Answer 1

The equations of the given lines are
$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$
$\therefore x_1=1,y_1=-1,z_1=1,x_2=3,y_2=k,z_2=0$
$l_1=2,m_1=3,n_1=4,l_2=1,m_2=2,n_2=1$
Since these lines intersect each other , we get
$\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ l_1& m_1& n_1\\ l_2& m_2& n_2\end{array}\right|=0$
$\left|\begin{array}{ccc}2& k+1& -1\\ 2& 3& 4\\ 1& 2& 1\end{array}\right|=0$
$\Rightarrow$ $2(3-8)-(k+1)(2-4)-1(4-3)=0$
$\Rightarrow$ $-10+2k+2-1=0$
$\Rightarrow$ $2k=9$ $\Rightarrow$ $k=\frac{9}{2}$