
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
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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.
(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.
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