The equation of the plane passing through the line of intersection of the planes x y z 1 2x 3y 4z 7 ...........

Mathematics (NDA) Coordinate Geometry

The equation of the plane passing through the line of intersection of the planes x + y + z = 1, 2x + 3y + 4z = 7, and perpendicular to the plane x – 5y + 3z = 5 is given by

x + 2y + 3z – 6 = 0

x + 2y + 3z + 6 = 0

3x + 4y + 5z – 8 = 0

3x + 4y + 5z + 8 = 0

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Explanation:

Let P1 : x + y + z – 1= 0
P2 : 2x + 3y + 4z – 7 = 0
Equation of plane passing through the line of intersection of P1 and P2 is given by
x + y + z – 1 + $\lambda$ (2x + 3y + 4z – 7) = 0
x(1 + 2 $\lambda$ ) + y(1 + 3 $\lambda$ ) + z(1 + 4 $\lambda$ ) – 1 – 7 $\lambda$ = 0
This is perpendicular to x –5y –3z –5 = 0
=> 1(1 + 2 $\lambda$ ) – 5(1 + 3 $\lambda$ ) + 3(1 + 4 $\lambda$ ) = 0
=>1 + 2 $\lambda$ –5 – 15 $\lambda$ + 3 + 12 $\lambda$ = 0
=> – $\lambda$ – 1 = 0 => $\lambda$ = – 1
$\therefore$ Equation of plane is
– x – 2y – 3z – 1 + 7 = 0 => x + 2y + 3z -6 = 0