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Question
If the lines \[\frac{x - 1}{- 3} = \frac{y - 2}{- 2k} = \frac{z - 3}{2} \text{ and }\frac{x - 1}{k} = \frac{y - 2}{1} = \frac{z - 3}{5}\] are perpendicular, find the value of k and, hence, find the equation of the plane containing these lines.
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Solution
\[\text{ We know that the lines } \]
\[\frac{x - x_1}{l_1} = \frac{y - y_1}{m_1} = \frac{z - z_1}{n_1} \text{ and }\frac{x - x_2}{l_2} = \frac{y - y_2}{m_2} = \frac{z - z_2}{n_2} \text{ are perpendicular if } \]
\[ l_1 l_2 + m_1 m_2 + n_1 n_2 = 0\]
\[\text{ Here } ,\]
\[ l_1 = - 3; m_1 = - 2k; n_1 = 2; l_2 = k; m_2 = 1; n_2 = 5\]
\[\text{ It is given that given lines are perpendicular } .\]
\[ \Rightarrow l_1 l_2 + m_1 m_2 + n_1 n_2 = 0\]
\[ \Rightarrow \left( - 3 \right) \left( k \right) + \left( - 2k \right) \left( 1 \right) + \left( 2 \right) \left( 5 \right) = 0\]
\[ \Rightarrow - 3k - 2k + 10 = 0\]
\[ \Rightarrow - 5k = - 10\]
\[ \Rightarrow k = 2\]
\[\text{ Substituting this value in the given equations of the lines, we get } \]
\[\frac{x - 1}{- 3} = \frac{y - 2}{- 4} = \frac{z - 3}{2} . . . \left( 1 \right)\]
\[ \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 3}{5} . . . \left( 2 \right)\]
\[\text{ Finding the equation of the plane } \]
\[\text{ Let the direction ratios of the required plane be proportional to a, b, c . } \]
\[\text{ We know from (1) and (2) that lines (1) and (2) pass through the point (1, 2, 3) and the direction ratios of (1) and (2) are proportional to -3, -4, 2 and 2, 1, 5 respectively.} \]
\[\text{ Since the plane contains the lines (1) and (2), the plane must pass through the point (1, 2, 3) and it must be parallel to the line. } \]
\[\text{ So, the equation of the plane is } \]
\[a \left( x - 1 \right) + b \left( y - 2 \right) + c \left( z - 3 \right) = 0 . . . \left( 3 \right)\]
\[ - 3a - 4b + 2c = 0 . . . \left( 4 \right)\]
\[2a + b + 5c = 0 . . . \left( 5 \right)\]
\[\text{ Solving (1), (2) and (3), we get} \]
\[\begin{vmatrix}x - 1 & y - 2 & z - 3 \\ - 3 & - 4 & 2 \\ 2 & 1 & 5\end{vmatrix} = 0\]
\[ \Rightarrow - 22 \left( x - 1 \right) + 19 \left( y - 2 \right) + 5 \left( z - 3 \right) = 0\]
\[ \Rightarrow - 22x + 19y + 5z = 31\]
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