Question

Find the Cartesian equation of the plane, passing through the line of intersection of the planes `vecr. (2hati + 3hatj - 4hatk) + 5 = 0`and `vecr. (hati - 5hatj + 7hatk) + 2 = 0`  intersecting the y-axis at (0, 3).

Answer 1

The given planes are :

`vecr. (2hati + 3hatj - 4hatk) + 5 = 0`

`vecr. (hati - 5hatj + 7hatk) + 2 = 0`

Any plane through the line of intersection of the given planes is :

`vecr. { (2hati + 3hatj - 4hatk) + λ  (hati - 5hatj + 7hatk) } = -5 - λ(2)`      ...(1)

It passes through the point (0, 3, 0), i.e;

`(0hati + 3hatj + 0hatk) . { (2hati + 3hatj - 4hatk) + λ (hati - 5hatj + 7hatk) } = - 5 -2 λ`

`3hatj . { (2 + λ) hati + (3 - 5 λ) hatj + (-4 + 7 λ) hatk } = - 5 -2 λ`

⇒ 3(3 - 5λ) = -5 - 2λ
⇒ 9 - 15λ = -5 - 2λ
⇒ 14  = 15λ - 2λ
⇒  13λ = 14

λ = `(14)/(13)`

Substituting this value of λ in (1), we have

`vecr. { (2hati + 3hatj - 4hatk) +(14)/(13) (hati - 5hatj + 7hatk )} = -5 - 2  (14/13)`

⇒ `vecr. (40hati - 31hatj + 46hatk) = -65 - 28`

⇒ `vecr. (40hati - 31hatj + 46hatk) = -93`

⇒ `vecr. (40hati - 31hatj + 46hatk) + 93 = 0`

Which is the required equation.