The vector equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y+ 4z = 5 which is perpendicular to the plane x – y + z = 0 is :

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Answer: C. \mathop r\limits^ 	o . \left( {\mathop i\limits^ \wedge - \mathop k\limits^ \wedge } ight) + 2 = 0 Given, P1 : x + y + z = 1 P1 : 2x + 3y + 4z = 5 Equation of the plane passing through the line of intersection of the plane P1 and P2 is : P1 + \lambda P2 = 0 \Rightarrow (x + y + z –1) + \lambda (2x + 3y + 4z – 5) = 0 \Rightarrow x(1 + 2\lambda ) + y(1 + 3\lambda ) + z(1 + 4\lambda ) - 5\lambda - 1 = 0 .....(1) Direction Ratio (D.R) of this plane = (1 + 2\lambda , 1 + 3\lambda , 1 + 4\lambda ) Plane (1) is perpendicular to x - y + z = 0, whose D.R = (1, -1, 1) As they are perpendicular so dot product of D.R = 0 	herefore (1) (1 + 2\lambda ) + (–1) (1 + 3\lambda ) + (1) (1 + 4\lambda ) = 0 \Rightarrow 1 + 2\lambda –1 – 3\lambda + 1 + 4\lambda = 0 \Rightarrow \lambda = - {1 \over 3} Putting the value of \lambda in equation (1), we get \Rightarrow {x \over 3} - {z \over 3} + {2 \over 3} = 0 \Rightarrow x - z + 2 = 0 Vector form of this plane, \mathop r\limits^ 	o . \left( {\mathop i\limits^ \wedge - \mathop k\limits^ \wedge } ight) + 2 = 0

The vector equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y+ 4z = 5 which is perpendicular to the plane x – y + z = 0 is :

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