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

Answer: (A)

$K=0$

Two lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ are coplanar if

$\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=0$

where $x_1=2,y_1=3,z_1=4,x_2=1,y_2=4,z_2=5$ and $a_1=1,b_1=1,c_1=-K,a_2=K,b_2=2,c_2=1$

$\Rightarrow \left|\begin{array}{ccc}1-2& 4-3& 5-4\\ 1& 1& -K\\ K& 2& 1\end{array}\right|=0$

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

$\Rightarrow -1(1+2K)-1(1+K^2)+1(2-K)=0$

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

$\Rightarrow -K^2-3K=0$

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

$\therefore K=0,-3$

For $k=0$ we have

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

$\Rightarrow -1(1-0)-1(1-0)+1(2-0)=-1-1+2=0$

For $k=-3$ we have

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

$\Rightarrow -1(1-6)-1(1+9)+1(2+3)=0$

$\Rightarrow 5-10+5=0$

$\therefore k=-3$

Hence $k=0,-3$ are valid.

Thus, option $A$ only