Question

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

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

Answer 1

The correct option is B

$k=0$

Explanation of the correct option:

Finding the value of k:

Compare $\frac{\left[x-2\right]}{1}=\frac{\left[y-3\right]}{1}=\frac{\left[z-4\right]}{-k}$ with $\frac{\left[x-x_1\right]}{{\mathbf{a}}_{\mathbf{1}}}\mathbf{=}\frac{\left[y-y_1\right]}{{\mathbf{b}}_{\mathbf{1}}}\mathbf{=}\frac{\left[z-z_1\right]}{{\mathbf{c}}_{\mathbf{1}}}$.

Therefore, $x_1=2,y_1=3,z_1=4,a_1=1,b_1=1,c_1=-k$.

Similarly, compare $\frac{\left[x-1\right]}{k}=\frac{\left[y-4\right]}{2}=\frac{\left[z-5\right]}{1}$with $\frac{\left[x-x_2\right]}{{\mathbf{a}}_{\mathbf{2}}}\mathbf{=}\frac{\left[y-y_2\right]}{{\mathbf{b}}_{\mathbf{2}}}\mathbf{=}\frac{\left[z-z_2\right]}{{\mathbf{c}}_{\mathbf{2}}}$.

Hence, $x_2=1,y_2=4,z_2=5,a_2=k,b_2=2,c_2=1$

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

$$\begin{gathered} \Rightarrow \left|\begin{array}{ccc}-1& 1& 1\\ 1& 1& -k\\ k& 2& 1\end{array}\right|=0 \\ \Rightarrow -(1+2k)-1(1+k^2)+1(2-k)=0 \\ \Rightarrow -1-2k-1-k^2+2-k=0 \\ \Rightarrow k^2+3k=0 \\ \Rightarrow k=0,k=-3 \end{gathered}$$

Hence, the correct answer is option (B)