Question

Cartesian Form of plane of three dimensional geometry.

Answer 1

Let equations of two planes

a₁x + b₁y + c₁z + d₁ = 0 ...(1)

and a₂x + b₂y + c₂z + d₂ = 0 ... (2)

In figure, plane PQ and RS intersect in line AB. Thus, AB is a intersecting line.

∵ Planes (1) and (2) intesect in a line, then we suppose that the point (x'₁ y'₁, z'₁) lies on this line. Because this point is on the intersect line, thus its coordinate satisfy (1) and (2).

a₁x' + b₁y’ + c₁z' + d₁ = 0 ...(3)

and a₂x' + b₂y' + c₂z' + d₂ = 0 ...(4)

Multiplying equation (4) by any constant X and adding in equation (3), we get

(a₁x' + b₁y' + C₁z' + d₁) + λ(a₂x' + b₂y' + c₂z' + d₂) = 0

⇒ (a₁ + λa₂)x' + (b₁ + λb₂)y' + (c₁ + λc₂)z' + d₁ + λd₂ = 0

∴ Required plane

(a₁ + λa₂)x + (b₁ + λb₂)y + (c₁ + λc₂)z + (d₁ + λd₂) = 0 Value of λ can be find by the given condition.