Determine the equation of the plane that passes through the point $(3, - 1, 2)$ and contains the line of intersection of the planes $2 x - y + z = 5$ and $x + y - 4 z = - 2$ .

hard 4 marks Class 12 · Mathematics · Vector Algebra

Answer

x - z = 1

The correct answer is $x - z = 1$ — Divide by 3 to obtain the simplest form: $x - z - 1 = 0$ , or $x - z = 1$ .

Step-by-step solution

Write the two given planes as $P_{1} : 2 x - y + z - 5 = 0$ and $P_{2} : x + y - 4 z + 2 = 0$ .
A plane containing their line of intersection can be expressed as $P = P_{1} + λ P_{2} = 0$ , i.e.,
$(2 + λ) x + (- 1 + λ) y + (1 - 4 λ) z + (- 5 + 2 λ) = 0$ .
Substitute the point $(3, - 1, 2)$ into this family to find $λ$ :
$(2 + λ) \cdot 3 + (- 1 + λ) (- 1) + (1 - 4 λ) \cdot 2 + (- 5 + 2 λ) = 0$ ,
Simplify: $6 + 3 λ + 1 - λ + 2 - 8 λ - 5 + 2 λ = 0 \Rightarrow 4 - 4 λ = 0$ ,
Thus $λ = 1$ .
Insert $λ = 1$ back into the family:
$(2 + 1) x + (- 1 + 1) y + (1 - 4) z + (- 5 + 2) = 0 \Rightarrow 3 x + 0 y - 3 z - 3 = 0$ .
Divide by 3 to obtain the simplest form: $x - z - 1 = 0$ , or $x - z = 1$ .

Common mistake: Forgetting to include the constant terms when forming the combined plane equation.

Concept tested: Plane through line of intersection of two planes