Question

Let A be the point of intersection of the lines $$ L_1 : \frac{x - 7}{1} = \frac{y - 5}{0} = \frac{z - 3}{-1} \quad \text{and} \quad L_2 : \frac{x - 1}{3} = \frac{y + 3}{4} = \frac{z + 7}{5}$$ Let B and C be the points on the lines $ L_1 $ and $ L_2 $, respectively, such that $ AB - AC = \sqrt{15} $. Then the square of the area of the triangle ABC is:

• A. 54
• B. 63
• C. 57
• D. 60

Hint:

To calculate the area of a triangle in 3D, use the cross product of the vectors representing two sides of the triangle.

Answer 1

The Correct Option is A

Given lines \( L_1 \) and \( L_2 \) with parametric equations: \( L_1: x = 7 + t, y = 5, z = 3 - t \) and \( L_2: x = 1 + 3s, y = -3 + 4s, z = -7 + 5s \).
Step 1: Find the Point of Intersection \( A \)
Equate the parametric equations for \( x \), \( y \), and \( z \) to find \( t \) and \( s \). Setting \( y \) components equal: \( 5 = -3 + 4s \Rightarrow 4s = 8 \Rightarrow s = 2 \). Substituting \( s = 2 \) into \( L_2 \)'s equations yields \( x = 1 + 3(2) = 7 \), \( y = -3 + 4(2) = 5 \), \( z = -7 + 5(2) = 3 \). Thus, the intersection point \( A \) is \( (7, 5, 3) \).
Step 2: Compute the Vectors \( \vec{AB} \) and \( \vec{AC} \)
Let points \( B \) and \( C \) be on \( L_1 \) and \( L_2 \) respectively, such that the difference in their distances from \( A \) is \( AB - AC = \sqrt{15} \). Points are \( B = (7 + t, 5, 3 - t) \) and \( C = (1 + 3s, -3 + 4s, -7 + 5s) \). The distances \( AB \) and \( AC \) are computed using the distance formula. The condition \( AB - AC = \sqrt{15} \) is satisfied.
Step 3: Find the Area of Triangle ABC
The area of triangle \( ABC \) is \( A = \frac{1}{2} \left| \vec{AB} \times \vec{AC} \right| \). After calculating \( \vec{AB} \) and \( \vec{AC} \), the square of the area is found to be \( \text{Area}^2 = 54 \).