Question

Find the equation of the plane containing the line `x/(-2) = (y - 1)/3 = (1 - z)/1` and the point (–1, 0, 2).

Answer 1

Let the equation of the plane be a(x – x₁) + b(y – y₁) + c(z – z₁) = 0, which is passing through (–1, 0, 2).

⇒ a(x + 1) + b(y – 0) + c(z – 2) = 0   ...(1)

Given the line `(x - 0)/(-2) = (y - 1)/3 = (z - 1)/(-1)` passing through (0, 1, 1) and having direction ratios (−2, 3, −1).

Since the plane contains the line and the point (0, 1, 1),

⇒ a(0 + 1) + b(1 − 0) + c(1 – 2) = 0

⇒ a + b – c = 0  ...(2)

Also the line and normal to the plane are perpendicular

⇒ (a, b, c) × (−2, 3, −1) = 0

⇒ –2𝑎 + 3b – c = 0  ...(3)

Solving (2) and (3)

From (2),

c = a + b

Put in (3),

−2a + 3b − (a + b) = 0

⇒ −3a + 2b = 0

⇒ 2b = 3a

⇒ b = `(3a)/2`

Then, c = a + b

= `a + (3a)/2`

= `(5a)/2`

⇒ a : b : c = 2 : 3 : 5

⇒ `a/2 = b/3 = c/5 = k`

Hence, the required equation of the plane is

⇒ 2(x + 1) + 3(y − 0) + 5(z − 2) = 0

⇒ 2x + 2 + 3y + 5z − 10 = 0 

⇒ 2x + 3y + 5z − 8 = 0

Answer 2

Let the equation of the plane be a(x – x₁) + b(y – y₁) + c(z – z₁) = 0, which is passing through (–1, 0, 2).

⇒ a(x + 1) + b(y – 0) + c(z – 2) = 0   ...(1)

Given the line `(x - 0)/(-2) = (y - 1)/3 = (z - 1)/(-1)` passing through (0, 1, 1) and having direction ratios (−2, 3, −1).

Since the plane contains the line and the point (0, 1, 1),

⇒ a(0 + 1) + b(1 − 0) + c(1 – 2) = 0

⇒ a + b – c = 0  ...(2)

Also the line and normal to the plane are perpendicular

⇒ (a, b, c) × (−2, 3, −1) = 0

⇒ –2𝑎 + 3b – c = 0  ...(3)

Hence, the required equation of the plane is

`|(x + 1, y, z - 2),(1, 1, -1),(-2, 3, -1)| = 0`

`(x + 1) |(1,-1),(3,-1)| - y|(1,-1),(-2,-1)| + (z - 2) |(1,1),(-2,3)| = 0`

⇒ (x + 1)[(1) (−1) − (−1) (3)] − y[(1) (−1) − (−1) (−2)] + (z − 2)[(1) (3) − (1) (−2)] = 0

⇒ (x + 1)[−1 + 3] − y[−1 − 2] + (z − 2)[3 + 2] = 0

⇒ 2(x + 1) + 3y + 5(z − 2) = 0

⇒ 2x + 2 + 3y + 5z − 10 = 0

⇒ 2x + 3y + 5z − 8 = 0