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 given lines are: 

Line \( L_1: \frac{x + 2}{1} = \frac{y - 1}{1} = \frac{z}{1} = \lambda \), where any point on it is \( (\lambda - 2, \lambda + 1, \lambda) \)

Line \( L_2: \frac{x - 3}{5} = \frac{y - 1}{-1} = \frac{z - 1}{1} = \mu \), where any point on it is \( (5\mu + 3, -\mu, \mu + 1) \)

Line \( L_3: \frac{x}{-3} = \frac{y - 3}{3} = \frac{z - 2}{1} = k \), where any point on it is \( (-3k, 3k + 3, k + 2) \)

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

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

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

Now, let's calculate the vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \):

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

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

The area \( A \) is given by:

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

Computing the cross product:

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

Thus, the square of the area is:

\( A^2 = 56 \)

Answer 2

We are given the parametric equations of two lines.
Let's first express these lines in vector form. 
Step 1: Parametric equations of the lines For the first line \( \frac{x + 2}{-3} = \frac{y - 3}{3} = \frac{z - 2}{1} \), we can express it as: \[ (x, y, z) = (-2, 3, 2) + t(-3, 3, 1) \] where \( t \) is a parameter. For the second line \( \frac{x - 3}{5} = \frac{y}{-1} = \frac{z - 1}{1} \), we can express it as: \[ (x, y, z) = (3, 0, 1) + s(5, -1, 1) \] where \( s \) is a parameter. 
Step 2: Finding the vectors representing the lines The direction vector of the first line is \( \vec{v_1} = (-3, 3, 1) \), and the direction vector of the second line is \( \vec{v_2} = (5, -1, 1) \). 
Step 3: Finding the cross product of the direction vectors The area of the triangle formed by the two lines and the origin can be calculated using the formula for the area of a triangle formed by two vectors: \[ A = \frac{1}{2} |\vec{v_1} \times \vec{v_2}| \] Now, calculate the cross product \( \vec{v_1} \times \vec{v_2} \): \[ \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\-3 & 3 & 1 \\5 & -1 & 1 \end{vmatrix} \] Expanding the determinant: \[ \vec{v_1} \times \vec{v_2} = \hat{i} \begin{vmatrix} 3 & 1 \\-1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} -3 & 1 \\5 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} -3 & 3 \\5 & -1 \end{vmatrix} \] \[ = \hat{i}(3 \cdot 1 - (-1) \cdot 1) - \hat{j}(-3 \cdot 1 - 5 \cdot 1) + \hat{k}(-3 \cdot (-1) - 3 \cdot 5) \] \[ = \hat{i}(3 + 1) - \hat{j}(-3 - 5) + \hat{k}(3 - 15) \] \[ = 4\hat{i} + 8\hat{j} - 12\hat{k} \] 
Step 4: Finding the magnitude of the cross product Now, calculate the magnitude of \( \vec{v_1} \times \vec{v_2} \): \[ |\vec{v_1} \times \vec{v_2}| = \sqrt{4^2 + 8^2 + (-12)^2} = \sqrt{16 + 64 + 144} = \sqrt{224} = 2\sqrt{56} \] 
Step 5: Finding the area of the triangle The area of the triangle is: \[ A = \frac{1}{2} |\vec{v_1} \times \vec{v_2}| = \frac{1}{2} \times 2\sqrt{56} = \sqrt{56} \] Thus, \( A^2 = 56 \).