Question

Let P₁ and P₂ be two planes given by P₁ : 10x + 15y + 12z - 60 = 0 , P₂ : -2x + 5y + 4z - 20 = 0. Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on P₁ and P₂ ?

(A) \(\frac{x-1}{0}=\frac{y-1}{0}=\frac{z-1}{5}\)

(B) \(\frac{x-6}{-5}=\frac{y}{2}=\frac{z}{3}\)

(C) \(\frac{x}{-2}=\frac{y-4}{5}=\frac{z}{4}\)

(D) \(\frac{x}{1}=\frac{y-4}{-2}=\frac{z}{3}\)

Answer 1

(A) \(\frac{x-1}{0}=\frac{y-1}{0}=\frac{z-1}{5}\)

(B) \(\frac{x-6}{-5}=\frac{y}{2}=\frac{z}{3}\)

(D) \(\frac{x}{1}=\frac{y-4}{-2}=\frac{z}{3}\)

The line should be either coincident on P₁ or on P₂ or intersect on P₁ and P₂ on different points.

  (D) \(\frac{x}{1}=\frac{y-4}{-2}=\frac{z}{3}\)

⇒ (λ, -2λ + 4,3λ) lie on P₂

 (A) \(\frac{x-1}{0}=\frac{y-1}{0}=\frac{z-1}{5}\) intersects P₁ and P₂ on different points.

 (B) \(\frac{x-6}{-5}=\frac{y}{2}=\frac{z}{3}\) also intersects P₁ and P₂ on different points.