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

We are given two lines \( L_1 \) and \( L_2 \) with parametric equations: - For \( L_1 \), since \( \frac{y - 5}{0} \) implies \( y = 5 \), we can parametrize \( L_1 \) as: \[ x = 7 + t, \quad y = 5, \quad z = 3 - t \] - For \( L_2 \), the parametric equations are: \[ x = 1 + 3s, \quad y = -3 + 4s, \quad z = -7 + 5s \] 
Step 1: Find the Point of Intersection \( A \)
To find the point of intersection, solve for \( t \) and \( s \) by equating the parametric equations for \( x \), \( y \), and \( z \). 
- From \( y \), we already know \( y = 5 \) for \( L_1 \). 
So for \( L_2 \), set \( y = -3 + 4s = 5 \): \[ -3 + 4s = 5 \quad \Rightarrow \quad 4s = 8 \quad \Rightarrow \quad s = 2 \] 
- Now, substitute \( s = 2 \) into the parametric equations of \( L_2 \): \[ x = 1 + 3(2) = 7, \quad y = -3 + 4(2) = 5, \quad z = -7 + 5(2) = 3 \] 
Thus, the point of intersection \( A \) is \( (7, 5, 3) \). 
Step 2: Compute the Vectors \( AB \) and \( AC \)
Let the points \( B \) and \( C \) be points on lines \( L_1 \) and \( L_2 \) such that \( AB - AC = \sqrt{15} \). 
Using the parametric equations of \( L_1 \) and \( L_2 \), we find the coordinates of \( B \) and \( C \). - \( B = (7 + t, 5, 3 - t) \) - \( C = (1 + 3s, -3 + 4s, -7 + 5s) \) Using the distance formula, we compute the distances \( AB \) and \( AC \). After solving, we find that \( AB - AC = \sqrt{15} \). 
Step 3: Find the Area of Triangle ABC
The area of triangle \( ABC \) is given by the magnitude of the cross product of vectors \( \vec{AB} \) and \( \vec{AC} \): \[ A = \frac{1}{2} \left| \vec{AB} \times \vec{AC} \right| \] After calculating the vectors \( \vec{AB} \) and \( \vec{AC} \), we find that the square of the area is: \[ \text{Area}^2 = 54 \] 
Thus, the square of the area of the triangle is \( 54 \).

Answer 2

Step 1: The lines \( L_1 \) and \( L_2 \) are given as: 

\[ L_1: \frac{x - 7}{1} = \frac{y - 5}{0} = \frac{z - 3}{-1} \] \[ L_2: \frac{x - 1}{3} = \frac{y + 3}{4} = \frac{z + 7}{5} \]

Step 2: The cosine of the angle \( \theta \) is given by:

\[ \cos \theta = \left| \frac{3 + 0 - 5}{\sqrt{2} \times \sqrt{50}} \right| \] Simplifying: \[ \cos \theta = \frac{2}{10} = \frac{1}{5} \]

Step 3: Using the identity \( \sin^2 \theta = 1 - \cos^2 \theta \), we get:

\[ \sin \theta = \frac{2\sqrt{6}}{5} \]

Step 4: The area of the triangle is given by:

\[ \text{Area} = \frac{1}{2} \times ab \times \sin \theta \] Substituting the known values: \[ \text{Area} = \frac{1}{2} \times \sqrt{15} \times \sqrt{15} \times \frac{2\sqrt{6}}{5} \] Simplifying: \[ \text{Area} = 3\sqrt{6} \]

Step 5: The square of the area is:

\[ (\text{Area})^2 = 9 \times 6 = 54 \]