Question

find the equation of the plane through the line of intersection of the Planes X + Y + Z = 1 and 2 X + 3 Y + 4z =5 which is perpendicular to the plane x _y + Z = 0 also find the distance of the plane obtained above from the origin.

Answer 1

HEY !!

Answer:

= √2 units.

Step-by-step explanation:

The equation of a plane passing through the intersection of the given plane is

        ( x + y + z - 1 ) + λ (2x + 3y + 4z - 5) = 0

=>     ( 1 + 2λ )x + ( 1 + 3λ ) y + ( 1 + 4λ ) z - ( 1 + 5λ ) = 0     --------> (1)

Since, equation (1) is perpendicular x - y + z = 0

=> ( 1  + 2λ ) 1 + ( 1 + 3λ ) (-1) + ( 1 + 4λ ) 1 = 0

=>  1 + 2λ - 1 - 3λ + 1 + 4λ = 0

=> 3λ + 1 = 0

=> λ = $-\frac{1}{3}$

Putting the value of λ in (1) we get,

    $(1-\frac{2}{3} )x + ( 1 - 1 )y + (1-\frac{4}{3})z - (1-\frac{5}{3} ) = 0$

=> $\frac{x}{3} - \frac{z}{3} + \frac{2}{3} = 0$

=> x - z + 2 = 0 it is required plane.

Let d be the distance of this plane from origin.

∴             d = | 0. x+ 0. y + 0. (-z) + 2 / √(1)² + (0)² + (-1)² |

    => | 2/√2 |

FINAL RESULT = 2√2 Units.

GOOD LUCK !