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 $k$ is equal to

• A. $-1$
• B. $\frac{2}{9}$
• C. $\frac{9}{2}$
• D. $0$

Answer 1

The correct option is C $\frac{9}{2}$

Here we can take four vectors

For line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$:
$\overrightarrow{a}=\hat{i}-j+k$
$\overrightarrow{b}=2i+3j+4k$

For line $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$:
$\overrightarrow{c}=3i+kj$
$\overrightarrow{d}=i+2j+k$
If these lines intersect, then $(\overrightarrow{a}-\overrightarrow{c}),\overrightarrow{b}$and$\overrightarrow{d}$ should be coplanar

$\therefore \left[\right(\overrightarrow{a}-\overrightarrow{c}),\overrightarrow{b},\overrightarrow{d}]=0$

$\left|\begin{array}{ccc}2& k+1& -1\\ 2& 3& 4\\ 1& 2& 1\end{array}\right|=0$

$\Rightarrow 2(-5)-(k+1)(-2)+(-1)\left(1\right)=0$
$\Rightarrow (k+1)\left(2\right)=11$
$\Rightarrow k=\frac{9}{2}$