Question

Let the area of the triangle formed by the lines $ \frac{x + 2}{-3} = \frac{y - 3}{3} = \frac{z - 2}{1} $, $ \frac{x - 3}{5} = \frac{y}{-1} = \frac{z - 1}{1} $ be $ A $. Then $ A^2 $ is equal to:

Hint:

When finding the area of a triangle formed by vectors, calculate the magnitude of the cross product of the direction vectors and divide by 2. The area squared can then be found easily.

Answer 1

Correct Answer: 56

The provided lines are:

Line \( L_1 \): \( \frac{x + 2}{1} = \frac{y - 1}{1} = \frac{z}{1} = \lambda \). A generic point on \( L_1 \) is \( (\lambda - 2, \lambda + 1, \lambda) \).

Line \( L_2 \): \( \frac{x - 3}{5} = \frac{y - 1}{-1} = \frac{z - 1}{1} = \mu \). A generic point on \( L_2 \) is \( (5\mu + 3, -\mu, \mu + 1) \).

Line \( L_3 \): \( \frac{x}{-3} = \frac{y - 3}{3} = \frac{z - 2}{1} = k \). A generic point on \( L_3 \) is \( (-3k, 3k + 3, k + 2) \).

Point \( P \), the intersection of \( L_1 \) and \( L_2 \), is \( P = (-2, 1, 0) \).

Point \( Q \), the intersection of \( L_1 \) and \( L_3 \), is \( Q = (0, 3, 2) \).

Point \( R \), the intersection of \( L_2 \) and \( L_3 \), is \( R = (3, 0, 1) \).

Vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \) are calculated as follows:

\( \overrightarrow{PQ} = 2\hat{i} + 2\hat{j} + 2\hat{k} \)

\( \overrightarrow{PR} = 5\hat{i} - \hat{j} + \hat{k} \)

The area \( A \) is determined by the formula:

\( A = \frac{1}{2} | \overrightarrow{PQ} \times \overrightarrow{PR} | \)

The cross product is computed as:

\( \overrightarrow{PQ} \times \overrightarrow{PR} = \hat{i} \hat{j} \hat{k} \left| \begin{matrix} 2 & 2 & 2 \\ 5 & -1 & 1 \end{matrix} \right| = \sqrt{56} \)

Consequently, the square of the area is:

\( A^2 = 56 \)