Question

Find the equation of the plane passing through the line of intersection of the planes `vecr(hati + hatj + hatk)` = 10 and `vecr.(2hati + 3hatj - hatk)` + 4 = 0 and passing through (–2, 3, 1).

Answer 1

Given planes are: `vecr(hati + hatj + hatk)` = 10 and `vecr.(2hati + 3hatj - hatk)` + 4 = 0

Changing to cartesian form, the equations of these planes are x + y + z – 10 = 0 and 2x + 3y – z + 4 = 0

Equation of any plane through the intersection of these planes is (x + y + z – 10) + λ(2x + 3y – z + 4) = 0  ...(i)

Now, it passes through the point (–2, 3, 1)

∴ (–2 + 3 + 1 – 10) + λ[2 × (–2) + 3(3) – 1 + 4] = 0

⇒ –8 + λ(–4 + 9 + 3) = 0

⇒ –8 + 8λ = 0

⇒ 8λ = 8

λ = 1

On substituting the value of λ in equation (i), we get

(x + y + z – 10) + 1(2x + 3y – z + 4) = 0

⇒ 3x + 4y – 6 = 0, which is the required equation of plane in cartesian form.

In vector form, required equation of plane is: `vecr.(3hati + 4hatj) - 6` = 0