Question

What is the equation of the plane through \[\left( {2,3,4} \right)\] and parallel to the plane \[x + 2y + 4z = 5\]?
A. \[x + 2y + 4z = 10\]
B. \[x + 2y + 4z = 3\]
C. \[x + y + 2z = 2\]
D. \[x + 2y + 4z = 24\]

Answer 1

Hint: First, calculate the equation of the plane passing through the point \[\left( {2,3,4} \right)\]. Then calculate the direction ratios of the required plane on the basis of the given plane. In the end, substitute the values in the equation of the required plane and solve it to get the required answer.

Formula used: The equation of a plane passing through the point \[\left( {{x_1},{y_1},{z_1}} \right)\] with direction ratios \[\left( {a,b,c} \right)\]: \[a\left( {x - {x_1}} \right) + b\left( {y - {y_1}} \right) + c\left( {z - {z_1}} \right) = 0\]
The planes \[{a_1}x + {b_1}y + {c_1}z = {d_1}\] and \[{a_2}x + {b_2}y + {c_2}z = {d_2}\] are parallel when the ratios of each pair of coefficients are equal. Means \[\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}} = k\].

Complete step by step solution: Given:
The plane passes through the point \[\left( {2,3,4} \right)\] and parallel to the plane \[x + 2y + 4z = 5\].

Let’s calculate the equation of the plane.
Consider, \[a,b,c\] are the direction ratios of the normal plane passing through the point \[\left( {2,3,4} \right)\].
Then, the equation of the required plane is:
\[a\left( {x - 2} \right) + b\left( {y - 3} \right) + c\left( {z - 4} \right) = 0\] \[.....\left( 1 \right)\]

We know that two planes \[{a_1}x + {b_1}y + {c_1}z = {d_1}\] and \[{a_2}x + {b_2}y + {c_2}z = {d_2}\] are parallel when the ratios of each pair of coefficients are equal. Means \[\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}} = k\].
So, for the given planes we get
\[\dfrac{a}{1} = \dfrac{b}{2} = \dfrac{c}{4} = k\]
\[ \Rightarrow a = k,b = 2k,c = 4k\]
Substitute these values in the equation \[\left( 1 \right)\].
 \[k\left( {x - 2} \right) + 2k\left( {y - 3} \right) + 4k\left( {z - 4} \right) = 0\]
\[ \Rightarrow k\left[ {\left( {x - 2} \right) + 2\left( {y - 3} \right) + 4\left( {z - 4} \right)} \right] = 0\]
\[ \Rightarrow \left( {x - 2} \right) + 2\left( {y - 3} \right) + 4\left( {z - 4} \right) = 0\]
\[ \Rightarrow x - 2 + 2y - 6 + 4z - 16 = 0\]
\[ \Rightarrow x + 2y + 4z - 24 = 0\]
\[ \Rightarrow x + 2y + 4z = 24\]
Thus, the equation of the required plane is \[x + 2y + 4z = 24\].

Thus, Option (D) is correct.

Note: Remember the following forms of the equation of a plane:
General equation of a plane: \[ax + by + cz + d = 0\] , where \[d \ne 0\]
Equation of plane whose intercepts are \[a,b,c\] is: \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\]
Equation of a plane passing through the point \[\left( {{x_1},{y_1},{z_1}} \right)\] with direction ratios \[\left( {a,b,c} \right)\]: \[a\left( {x - {x_1}} \right) + b\left( {y - {y_1}} \right) + c\left( {z - {z_1}} \right) = 0\]