Question

If the distance between the plane, \[23x - 10y - 2z + 48 = 0\] and the plane containing the lines \[\dfrac{{x + 1}}{2} = \dfrac{{y - 3}}{4} = \dfrac{{z + 1}}{3}\] and \[\dfrac{{x + 3}}{2} = \dfrac{{y + 2}}{6} = \dfrac{{z - 1}}{\lambda }\], \[\lambda \in R\] is equal to \[\dfrac{k}{{\sqrt {633} }}\]. What is the value \[k\].

Answer 1

Hint: First we will find the intersection point of the given straight lines. The point lies on the plane in which the lines lie. By using the distance formula, we will calculate the distance between the plane and the point.

Formula used:
Any point on a line \[\dfrac{{x - {x_1}}}{a} = \dfrac{{y - {y_1}}}{b} = \dfrac{{z - {z_1}}}{c} = p\left( {{\rm{say}}} \right)\] are \[\left( {ap + {x_1},bp + {y_1},cp + {z_1}} \right)\].
The distance between the plane \[ax + by + cz + d = 0\] and \[\left( {{x_1},{y_1},{z_1}} \right)\] is \[\left| {\dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\].

Complete step by step solution:
Given lines are \[\dfrac{{x + 1}}{2} = \dfrac{{y - 3}}{4} = \dfrac{{z + 1}}{3}\] and \[\dfrac{{x + 3}}{2} = \dfrac{{y + 2}}{6} = \dfrac{{z - 1}}{\lambda }\].
Since the direction ratios of the lines are not the same. So, the lines are not parallel. Thus, the lines intersect each other.
Rewrite the equation of lines.
\[\dfrac{{x + 1}}{2} = \dfrac{{y - 3}}{4} = \dfrac{{z + 1}}{3} = p\left( {{\rm{say}}} \right)\] ……(1)
\[\dfrac{{x + 3}}{2} = \dfrac{{y + 2}}{6} = \dfrac{{z - 1}}{\lambda } = q\left( {{\rm{say}}} \right)\] …….(2)
Calculate the value of \[x,y,z\] from equation (1)
\[\dfrac{{x + 1}}{2} = p\]
\[ \Rightarrow x = 2p - 1\]
\[\dfrac{{y - 3}}{4} = p\]
\[ \Rightarrow y = 4p + 3\]
\[\dfrac{{z + 1}}{3} = p\]
\[ \Rightarrow z = 3p - 1\]
Any point on the line (1) is \[\left( {2p - 1,4p + 3,3p - 1} \right)\].
Calculate the value of \[x,y,z\] from equation (2)
\[\dfrac{{x + 3}}{2} = q\]
\[ \Rightarrow x = 2q - 3\]
\[\dfrac{{y + 2}}{6} = q\]
\[ \Rightarrow y = 6q - 2\]
\[\dfrac{{z - 1}}{\lambda } = q\]
\[ \Rightarrow z = \lambda q + 1\]
Any point on the line (1) is \[\left( {2q - 3,6q - 2,\lambda q + 1} \right)\].
Assume the intersection point of line (1) and (2) is \[\left( {2p - 1,4p + 3,3p - 1} \right)\] and \[\left( {2q - 3,6q - 2,\lambda q + 1} \right)\].
Therefore,
\[2p - 1 = 2q - 3\]
\[ \Rightarrow 2p - 2q = - 2\]
\[ \Rightarrow p - q = - 1\]….(3)
\[4p + 3 = 6q - 2\]
 \[ \Rightarrow 4p - 6q = - 5\]….(4)
\[3p - 1 = \lambda q + 1\]….(5)
Now we will solve equation (3) and (4)
Multiply 4 with equation (3) and subtract it from equation (4)
\[\begin{array}{*{20}{c}}{4p}& - &{6q}& = &{ - 5}\\{4p}& - &{4q}& = &{ - 4}\\\hline{}& - &{2q}& = &{ - 1}\end{array}\]
\[ \Rightarrow q = \dfrac{1}{2}\]
Substitute the value of \[q\] in the equation (3)
\[p - \dfrac{1}{2} = - 1\]
\[ \Rightarrow p = - 1 + \dfrac{1}{2}\]
\[ \Rightarrow p = - \dfrac{1}{2}\]
Put the value of \[p\] and \[q\] in the point \[\left( {2p - 1,4p + 3,3p - 1} \right)\]
\[\left( {2p - 1,4p + 3,3p - 1} \right) = \left( {2 \cdot \left( { - \dfrac{1}{2}} \right) - 1,4 \cdot \left( { - \dfrac{1}{2}} \right) + 3,3 \cdot \left( { - \dfrac{1}{2}} \right) - 1} \right)\]
\[ \Rightarrow \left( {2p - 1,4p + 3,3p - 1} \right) = \left( { - 2,1, - \dfrac{5}{2}} \right)\]
Now we will calculate the distance between the point \[\left( { - 2,1, - \dfrac{5}{2}} \right)\] and the plane \[23x - 10y - 2z + 48 = 0\].
The distance is \[\left| {\dfrac{{23 \cdot \left( { - 2} \right) - 10 \cdot 1 - 2 \cdot \left( { - \dfrac{5}{2}} \right) + 48}}{{\sqrt {{{23}^2} + {{\left( { - 10} \right)}^2} + {{\left( { - 2} \right)}^2}} }}} \right|\]
\[ = \left| {\dfrac{{ - 46 - 10 + 5 + 48}}{{\sqrt {529 + 100 + 4} }}} \right|\]
\[ = \left| {\dfrac{{ - 3}}{{\sqrt {633} }}} \right|\] units.
Given that the distance between plane and \[\left( { - 2,1, - \dfrac{5}{2}} \right)\] is \[\dfrac{k}{{\sqrt {633} }}\].
Equate \[\left| {\dfrac{{ - 3}}{{\sqrt {633} }}} \right|\] with \[\dfrac{k}{{\sqrt {633} }}\] to calculate the value of \[k\].
\[\left| {\dfrac{{ - 3}}{{\sqrt {633} }}} \right| = \dfrac{k}{{\sqrt {633} }}\]
\[ \Rightarrow \dfrac{3}{{\sqrt {633} }} = \dfrac{k}{{\sqrt {633} }}\]
\[ \Rightarrow k = 3\]
Hence the value of \[k\] is 3.

Note: Many students often do a common mistake to solve the equation \[\left| {\dfrac{{ - 3}}{{\sqrt {633} }}} \right| = \dfrac{k}{{\sqrt {633} }}\]. They find the value of \[k\] as -3 which is incorrect. The distance is always positive. So, \[\left| {\dfrac{{ - 3}}{{\sqrt {633} }}} \right| = \dfrac{3}{{\sqrt {633} }}\]. The correct value of \[k\] is 3.