Question

If the 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, then k can have

• A. any value
• B. exactly one value
• C. exactly two values
• D. exactly three values

Answer 1

Answer: (C)

exactly two values

Answer 2

Condition for two lines are coplanar \hspace15mm $\begin{vmatrix} x_1 -x_2 & y_1 - y_2 & z_1 - z_2 \\\ 1 & m_1 & n_1 \\\ l_1 & m_2 & m_3 \\\ \end{vmatrix}= 0$ Where, $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are the points lie on lines (i) and (ii) respectively and $< l_1,m_1,n_1>$ and $< l_2,m_2,n_2 >$ are the direction cosines of the line (i) and line (ii), respectively. $\therefore \begin{vmatrix} 2-1 & 3 -4 & 4 - 5 \\\ 1 & 1 & -k \\\ k & 2 & 1 \\\ \end{vmatrix}=0$ $\Rightarrow \begin{vmatrix} 1 & -1 & \- 1 \\\ 1 & 1 & -k \\\ k & 2 & 1 \\\ \end{vmatrix}=0$ $\Rightarrow$ $1 (1+2k)+\ (1+k^2) - (2-k) = 0$ $\Rightarrow$ \hspace15mm $k^2 + 2k + k = 0$ $\Rightarrow$ \hspace20mm $k^2\ +\ 3k\ = 0$ $\Rightarrow$ \hspace30mm k = 0, - 3 If 0 appears in the denominator, then the correct way of representing the equation of straight line is \hspace15mm $\frac{x-2}{1} = \frac{y-3}{1}; z = 4$