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Question
Show that the lines \[\frac{x + 4}{3} = \frac{y + 6}{5} = \frac{z - 1}{- 2}\] and 3x − 2y + z + 5 = 0 = 2x + 3y + 4z − 4 intersect. Find the equation of the plane in which they lie and also their point of intersection.
Solution
\[\text{ The equation of the given line is } \]
\[\frac{x + 4}{3} = \frac{y + 6}{5} = \frac{z - 1}{- 2}\]
\[\text{ The coordinates of any point on this line are of the form } \]
\[\frac{x + 4}{3} = \frac{y + 6}{5} = \frac{z - 1}{- 2} = \lambda\]
\[ \Rightarrow x = 3\lambda - 4; y = 5\lambda - 6; z = - 2\lambda + 1\]
\[\text{ So, the coordinates of the point on the given line are } \left( 3\lambda - 4, 5\lambda - 6, - 2\lambda + 1 \right). \text{ Since this point lies on the plane } \]
\[ 3x - 2y + z + 5 = 0, \]
\[3 \left( 3\lambda - 4 \right) - 2 \left( 5\lambda - 6 \right) + \left( - 2\lambda + 1 \right) + 5 = 0\]
\[ \Rightarrow 9\lambda - 12 - 10\lambda + 12 - 2\lambda + 1 + 5 = 0\]
\[ \Rightarrow - 3\lambda + 6 = 0\]
\[ \Rightarrow \lambda = 2\]
\[\text{ So, the coordinates of the point are } \]
\[\left( 3\lambda - 4, 5\lambda - 6, - 2\lambda + 1 \right)\]
\[ = \left( 3 \left( 2 \right) - 4, 5 \left( 2 \right) - 6, - 2 \left( 2 \right) + 1 \right)\]
\[ = \left( 2, 4, - 3 \right)\]
\[\text{ Substituting this point in another plane equation } 2x + 3y + 4z - 4 = 0, \text{ we get } \]
\[2 \left( 2 \right) + 3 \left( 4 \right) + 4 \left( - 3 \right) - 4 = 0\]
\[ \Rightarrow 4 + 12 - 12 - 4 = 0\]
\[ \Rightarrow 0 = 0\]
\[\text{ So, the point (2, 4, -3) lies on another plane too. So, this is the point of intersection of both the lines } .\]
\[\text{ Finding the plane equation } \]
\[\text{ Let the direction ratios be proportional to a, b, c } . \]
\[\text{ Since the plane contains the line } \frac{x + 4}{3} = \frac{y + 6}{5} = \frac{z - 1}{- 2}, \text{ it must pass through the point (-4, -6, 1) and is parallel to this line.} \]
\[\text{ So, the equation of plane is} \]
\[a \left( x + 4 \right) + b \left( y + 6 \right) + c \left( z - 1 \right) = 0 . . . \left( 1 \right)\]
\[\text{ and } \]
\[3a + 5b - 2c = 0 . . . \left( 2 \right)\]
\[\text{ Since the given plane contains the planes 3x - 2y + z + 5 = 0 = 2x + 3y + 4z - 4 } , \]
\[3a - 2b + c = 0 . . . \left( 3 \right)\]
\[2a + 3b + 4z = 0 . . . \left( 4 \right)\]
\[\text{ Solving (3) and (4) using cross-multiplication, we get } \]
\[\frac{a}{- 11} = \frac{b}{- 10} = \frac{c}{13} . . . \left( 5 \right)\]
\[\text{ Using (1), (2) and (5), the equation of plane is} \]
\[\begin{vmatrix}x + 4 & y + 6 & z - 1 \\ 3 & 5 & - 2 \\ 11 & 10 & - 13\end{vmatrix} = 0\]
\[ \Rightarrow - 45 \left( x + 4 \right) + 17 \left( y + 6 \right) - 25 \left( z - 1 \right) = 0\]
\[ \Rightarrow 45 \left( x + 4 \right) - 17 \left( y + 6 \right) + 25 \left( z - 1 \right) = 0\]
\[ \Rightarrow 45x - 17y + 25z + 53 = 0\]
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