Question

Let $P_1$ and $P_2$ be two planes given by $P_1: 10 x+15 y+12 z-60=0,$ $P_2:-2 x+5 y+4 z-20=0$ Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on $P_1$ and $P_2$ ?

• 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

Answer: (A), (B), (D)

$\frac{x-1}{0}=\frac{y-1}{0}=\frac{z-1}{5}$

Answer 2

The equation representing a pair of planes is given by:

S: (10x + 15y + 12z – 60) (–2x + 5y + 4z – 20) = 0

To determine whether a line could be an edge of the given tetrahedron, we need to find the general points on each line and then solve it with the equation S. If the resulting equation yields more than one value for the variable λ, then the line may constitute an edge of the specified tetrahedron.

(A) The point is expressed as (1, 1, 5λ + 1).
Thus, we have the equation: (60λ – 23) (20λ – 17) = 0
$\lambda=\frac{23}{60}\ \text{and} \ \frac{17}{20}$
So, it can be an edge of the tetrahedron.

(B) The point is given by (–5λ + 6, 2λ, 3λ).
Thus, we have the equation: (16λ) (32λ – 32) = 0
This equation yields λ = 0 and λ = 1.
Therefore, it can be an edge of the tetrahedron.

(C) The point is represented as (–2λ, 5λ + 4, 4λ).
Hence, we obtain the equation: (103λ) (45λ) = 0
Upon solving, we find that λ = 0 only.
Consequently, it cannot be an edge of the tetrahedron.

(D) The point is described as (λ, –2λ + 4, 3λ).
This leads to the equation: (16λ) (0) = 0
As this equation holds true for any value of λ, it results in infinitely many solutions.
Thus, it can be an edge of the tetrahedron.