Question

Lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$ and $\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}$ are coplanar, if

• A. $k=0$
• B. $k=-1$
• C. $k=2$
• D. $k=3$

Answer 1

The correct option is A $k=0$

$(2,3,4)$ and $(1,4,5)$ line is same plane

$\therefore (2-1,3-4,4-5)\equiv (1,-1,-1)$

$\Rightarrow (1,-1,-1),(1,1,-k)$ and $(k,2,1)$ are coplanar vectors :-

$[{\overrightarrow{a}}_1,\overrightarrow{b},\overrightarrow{c}]=0$

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

$1(1+2k)+1(1+k^2)-1(2-k)=0$

$\Rightarrow 1+2k+k^2+1-2+k=0$

$\Rightarrow k^2+3k=0$

$\Rightarrow k(k+3)=0$

$\Rightarrow k=0$ or $k=-3$