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Question:

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:

r→×(i^+k^)+2=0

r→×(i^−k^)+2=0

r→.(i^−k^)−2;=0

r→.(i^−k^)+2=0

Solution:

Correct option is C. r→.(i^−k^)+2=0
Let the plane be (x+y+z-1) + λ(2x+3y+4z-5) = 0
⇒ (2λ+1)x + (3λ+1)y + (4λ+1)z - (5λ+1) = 0
⊥ to the plane x-y+z=0
⇒ λ = -1/3
⇒ the required plane is x - z + 2 = 0