Question -
Answer -
Let the equation ofthe plane that passes through the two-given planes
x + y + z = 1 and 2x +3y + 4z = 5 is
(x + y + z – 1)+ λ (2x + 3y + 4z – 5) = 0
(2λ + 1) x +(3λ + 1) y + (4λ + 1) z -1 – 5λ = 0…… (1)
So the direction ratioof the plane is (2λ + 1, 3λ + 1, 4λ + 1)
And direction ratio ofanother plane is (1, -1, 1)
Since, both the planesare ⊥
So by substituting ina1a2 + b1b2 + c1c2 =0
(2λ + 1 × 1)+ (3λ + 1 × (-1)) + (4λ + 1 × 1) = 0
2λ + 1 –3λ – 1 + 4λ + 1 = 0
3λ + 1 = 0
λ = -1/3
Substitute the valueof λ in equation (1) we get,
x – z + 2 = 0
∴ The required equationof the plane is x – z + 2 = 0