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The equation of the plane passing through \[\left( {1,1,1} \right)\] and \[\left( {1, - 1, - 1} \right)\], perpendicular to \[2x - y + z + 5 = 0\] is
A) \[2x + 5y + z - 8 = 0\]
B) \[x + y - z - 1 = 0\]
C) \[2x + 5y + z + 4 = 0\]
D) \[x - y + z - 1 = 0\]


Answer
VerifiedVerified
160.8k+ views
Hint: At first we take a plane passing through any one of the given point. Now since another point passing through the plane, so it must satisfies the equation and also since the taking plane perpendicular to the another plane, so we get two equation. Solving two equations, we get the value and put those value in the assuming equation, we get the required equation of plane.



Formula Used: Equation of a plane passing through the point \[\left( {{x_1},{y_1},{z_1}} \right)\] is \[a\left( {x - {x_1}} \right) + b\left( {y - {y_1}} \right) + c\left( {z - {z_1}} \right) = 0\]. Equation of the plane perpendicular to the plane \[{a_1}x + {b_1}y + {c_1}z + {d_1} = 0\] is \[{a_1}a + {b_1}b + {c_1}c = 0\].



Complete step by step solution:Let, equation of any plane passing through the point \[\left( {1,1,1} \right)\] be \[a\left( {x - 1} \right) + b\left( {y - 1} \right) + c\left( {z - 1} \right) = 0\]-----(1).
Now the plane (1) is also passing through the point \[\left( {1, - 1, - 1} \right)\], so it must satisfies the equation (1), therefore \[a\left( {1 - 1} \right) + b\left( { - 1 - 1} \right) + c\left( { - 1 - 1} \right) = 0\]
\[ \Rightarrow a \cdot 0 - 2b - 2c = 0\]
\[ \Rightarrow - 2b - 2c = 0\]
\[ \Rightarrow b + c = 0\]----(2).

Again the plane (1) is perpendicular to \[2x - y + z + 5 = 0\], hence \[2a - b + c = 0\]----(3).
Now solving (2) and (3), we get \[a = b = 1\] and \[c = - 1\]. Now substituting these values in (1), we get
\[1 \times \left( {x - 1} \right) + 1 \times \left( {y - 1} \right) - 1 \times \left( {z - 1} \right) = 0\]
\[ \Rightarrow x - 1 + y - 1 - z + 1 = 0\]
\[ \Rightarrow x + y - z - 1 = 0\].
Hence the equation of the plane passing through \[\left( {1,1,1} \right)\] and \[\left( {1, - 1, - 1} \right)\] , perpendicular to \[2x - y + z + 5 = 0\] is option (B)\[x + y - z - 1 = 0\].



Option ‘B’ is correct



Note: It must be in our mind that if a plane passing through the point \[\left( {1, - 1, - 1} \right)\] then its \[x = 1,y = - 1\] and \[z = - 1\]. Also solving (2) and (3) equation we get \[b = - c\] and \[a = b\]. Hence \[a = b = - c\], the we assuming, \[a = b = 1\], then \[c = - 1\].