Question

Find the equation of the plane that passes through the point \[\left( {1,{\rm{ }}2,{\rm{ }}3} \right)\] and parallel to the plane \[2x + 3y - 4z = 0\].
A. \[2x + 3y + 4z = 4\]
B. \[2x + 3y + 4z + 4 = 0\]
C. \[2x - 3y + 4z + 4 = 0\]
D. \[2x + 3y - 4z + 4 = 0\]

Answer 1

Hint: First, find the equation of the required plane on the basis of its parallel plane. Then, substitute the values of the coordinates of the given point in the required equation of the plane and solve it to find the value of the variable. In the end, substitute the value of the variable in the equation of the plane and get the required answer.



Formula Used:The equation of the plane parallel to the plane \[ax + by + cz = {d_1}\] is \[ax + by + cz + {d_2} = 0\].



Complete step by step solution:Given:
A plane passes through the point \[\left( {1,{\rm{ }}2,{\rm{ }}3} \right)\] and it is parallel to the plane \[2x + 3y - 4z = 0\].

We know that the equations of the plane parallel to the plane \[2x + 3y - 4z = 0\] are in the form \[2x + 3y - 4z + k = 0\].
Now we have to calculate the value of \[k\].
It is given that the plane passes through the point \[\left( {1,{\rm{ }}2,{\rm{ }}3} \right)\].
So, the point satisfies the equation of the plane.
Substitute the coordinates of the point in the equation of the required plane.
We get,
\[2\left( 1 \right) + 3\left( 2 \right) - 4\left( 3 \right) + k = 0\]
\[ \Rightarrow 2 + 6 - 12 + k = 0\]
\[ \Rightarrow - 4 + k = 0\]
\[ \Rightarrow k = 4\]
Now substitute \[k = 4\] in the equation of the plane.
Thus, the equation of the plane that passes through the point \[\left( {1,{\rm{ }}2,{\rm{ }}3} \right)\] and parallel to the plane \[2x + 3y - 4z = 0\] is \[2x + 3y - 4z + 4 = 0\].



Option ‘D’ is correct



Note: If two planes \[{a_1}x + {b_1}y + {c_1}z + {d_1} = 0\] and \[{a_2}x + {b_2}y + {c_2}z + {d_2} = 0\] are parallel to each other, then \[\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}\].