Question

If the straight lines \(\frac{x-1}{2}=\frac{y+1}{K} = \frac{z}{2}\) and \(\space20mm \frac{x+1}{5}=\frac{y+1}{2} = \frac{z}{k}\) are coplanar, then the plane(s) containing these two lines is/are

• A. y + 2z = - 1
• B. y + z = - 1
• C. y - 2 = - 1
• D. y - 2z = - 1

Answer 1

The Correct Option is C

PLAN If the straight lines are coplanar. They the should lie in same plane. 
Description of Situation If straight lines are coplanar. 
\(\Rightarrow \space20mm \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ \end{vmatrix}= 0 \)Since,  
and \(\space15mm \frac{x+1}{5}=\frac{y+1}{2} = \frac{z}{k}\) are coplanar. 
\(\Rightarrow \space5mm \begin{vmatrix} 2 & 0 & 0 \\ 2 & K & 2 \\ 5 & 2 & K \\ \end{vmatrix}= 0\)
\(\Rightarrow\) \(\space15mm K^2 = 4 \Rightarrow K = \pm 2\)
\(\therefore\) \(\space15mm n_1 = b_1 \times d_1 = 6j - 6k, for\ k = 2\)
\(\therefore\) \(\space15mm n_2 = b_2 \times d_2 = 14j + 14k, for\ k = -2\) 
So, equation of planes are \((r - a ).n_1 = 0\)
\(\Rightarrow\) \(\space15mm y - z = - 1\) and \((r - a ).n_2 = 0\)
\(\Rightarrow\) \(\space15mm y + z = - 1\)