Question

Find the equation of the plane passing through the intersection of the planes \[x + 2y + 3z + 4 = 0\] and \[4x + 3y + 2z + 1 = 0\] and the origin.

a) \[3x + 2y + z + 1 = 0\]
b) \[3x + 2y + z = 0\]
c) \[2x + 3y + z = 0\]
d) \[x + y + z = 0\]

Answer 1

Hint: First, we have to find the equation of the plane passing through the intersection of given plane in terms of \[\lambda \]. The required plane passes through the origin. Substitute
(0, 0, 0) in the obtained equation in order to find the value of \[\lambda \]. Then, substitute the value of \[\lambda \]in the obtained equation to get the answer.



Formula Used:The equation of the plane passing through the point of intersection of planes \[{a_1}x + {b_1}y + {c_1}z + {d_1} = 0\] and \[{a_2}x + {b_2}y + {c_2}z + {d_2} = 0\]is given by

\[{a_1}x + {b_1}y + {c_1}z + {d_1} + \lambda \left( {{a_2}x + {b_2}y + {c_2}z + {d_2}} \right) = 0\], where\[\lambda \]is a scalar.



Complete step by step solution:The equation of the plane passing through the point of intersection of planes \[x + 2y + 3z + 4 = 0\] and \[4x + 3y + 2z + 1 = 0\]is given by

\[x + 2y + 3z + 4 + \lambda \left( {4x + 3y + 2z + 1} \right) = 0\] ---(1)

It passes through origin.

So, equation (1) satisfies the point (0, 0, 0).

\[0 + 2(0) + 3(0) + 4 + \lambda \left( {4(0) + 3(0) + 2(0) + 1} \right) = 0\]

\[4 + \lambda = 0\]

\[\lambda = - 4\]

The required equation of the plane is obtained by substituting \[\lambda = - 4\]in equation (1).

\[x + 2y + 3z + 4 - 4\left( {4x + 3y + 2z + 1} \right) = 0\]

\[x + 2y + 3z + 4 - 16x - 12y - 8z - 4 = 0\]

\[ - 15x - 10y - 5z = 0\]

\[3x + 2y + z = 0\]



Option ‘B’ is correct



Note: While calculating the equation of the plane passing through the plane's point of intersection, the equations of given planes can be interchanged in the formula. It won’t affect the answer.