Question

Let \( P \) be the image of the point \( Q(7, -2, 5) \) in the line \( L: \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4} \), and let \( R(5, p, q) \) be a point on \( L \). Then the square of the area of \( \triangle PQR \) is:

Hint:

When solving problems involving the image of a point on a line, use parametric equations for the line and substitute the coordinates of the given point to find the image. Then use the area formula for the triangle to find the required value.

Answer 1

Step 1: Given Information

The point \( Q(7, -2, 5) \) is reflected over the line \( L \) which has the parametric equation:

\( \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4} \)

Let the image of the point \( Q \) be \( P \), and \( R(5, p, q) \) be a point on the line \( L \). We need to find the square of the area of \( \triangle PQR \).

Step 2: Parametric Equation of the Line \( L \)

The given line equation can be written in parametric form as: \[ x = 1 + 2t, \quad y = -1 + 3t, \quad z = 4t \] where \( t \) is the parameter.

Step 3: Find the Coordinates of \( R \) on the Line

From the parametric equations, the coordinates of the point \( R \) are: \[ R(5, p, q) \implies x = 5, \quad y = p, \quad z = q \] Substituting \( x = 5 \) into the equation \( x = 1 + 2t \), we get: \[ 5 = 1 + 2t \quad \Rightarrow \quad t = 2 \] Substituting \( t = 2 \) into the equations for \( y \) and \( z \), we find: \[ p = -1 + 3(2) = 5, \quad q = 4(2) = 8 \] Therefore, the coordinates of \( R \) are \( R(5, 5, 8) \).

Step 4: Find the Image Point \( P \) of \( Q(7, -2, 5) \)

To find the image point \( P \) of \( Q(7, -2, 5) \), we need to use the formula for the reflection of a point across a line. The formula for the reflection of point \( Q(x_1, y_1, z_1) \) across a line with parametric equations \( x = x_0 + at \), \( y = y_0 + bt \), and \( z = z_0 + ct \) is: \( P(x_2, y_2, z_2) = \left( x_1 - 2 \frac{(x_1 - x_0) \cdot a + (y_1 - y_0) \cdot b + (z_1 - z_0) \cdot c}{a^2 + b^2 + c^2} a, \, y_1 - 2 \frac{(x_1 - x_0) \cdot a + (y_1 - y_0) \cdot b + (z_1 - z_0) \cdot c}{a^2 + b^2 + c^2} b, \, z_1 - 2 \frac{(x_1 - x_0) \cdot a + (y_1 - y_0) \cdot b + (z_1 - z_0) \cdot c}{a^2 + b^2 + c^2} c \right) \) 
Here, the direction ratios of the line \( L \) are \( a = 2, b = 3, c = 4 \), and the coordinates of \( Q \) are \( Q(7, -2, 5) \). Using this formula, we find the coordinates of \( P \).

Step 5: Calculate the Area of Triangle \( \triangle PQR \)

The area of triangle \( \triangle PQR \) can be found using the determinant formula for the area of a triangle with vertices \( P(x_1, y_1, z_1) \), \( Q(x_2, y_2, z_2) \), and \( R(x_3, y_3, z_3) \). The formula is: \[ \text{Area} = \frac{1}{2} \left| \vec{PQ} \times \vec{PR} \right| \] where \( \vec{PQ} \) and \( \vec{PR} \) are the vectors from \( P \) to \( Q \) and from \( P \) to \( R \), respectively. After calculating the vectors \( \vec{PQ} \) and \( \vec{PR} \), and performing the cross product, we find the area. The square of this area is the final answer.

Conclusion

The square of the area of triangle \( \triangle PQR \) is 957.

Answer 2

Step 1: Identify the given elements.
- Point \( Q = (7, -2, 5) \)
- Line \( L \) represented parametrically by:
\[ x = 1 + 2t, \quad y = -1 + 3t, \quad z = 0 + 4t \] A point \( R = (5, p, q) \) lies on \( L \), so there exists \( t \) such that:
\[ 5 = 1 + 2t \implies t = 2 \] Then,
\[ p = -1 + 3 \times 2 = 5, \quad q = 4 \times 2 = 8 \] Thus, \[ R = (5, 5, 8) \]

Step 2: Find the image \( P \) of \( Q \) in line \( L \).
The image of \( Q \) in line \( L \) is the reflection of point \( Q \) about \( L \).
To find \( P \), first find the foot \( F \) of the perpendicular from \( Q \) to \( L \), then use:
\[ P = 2F - Q \]

Parametric point on \( L \): \[ (1 + 2s, -1 + 3s, 4s) \] Vector from \( Q \) to point on \( L \): \[ \overrightarrow{QF} = (1 + 2s - 7, -1 + 3s + 2, 4s - 5) = (-6 + 2s, 1 + 3s, 4s - 5) \] Direction vector of line \( L \): \[ \mathbf{d} = (2, 3, 4) \] Since \( F \) is foot of perpendicular, \(\overrightarrow{QF}\) is perpendicular to \( \mathbf{d} \): \[ \overrightarrow{QF} \cdot \mathbf{d} = 0 \] Calculate dot product:
\[ (-6 + 2s) \times 2 + (1 + 3s) \times 3 + (4s - 5) \times 4 = 0 \] Simplify:
\[ -12 + 4s + 3 + 9s + 16s - 20 = 0 \] \[ (4s + 9s + 16s) + (-12 + 3 - 20) = 0 \] \[ 29s - 29 = 0 \Rightarrow s = 1 \]

Coordinates of foot \( F \): \[ x_F = 1 + 2 \times 1 = 3 \] \[ y_F = -1 + 3 \times 1 = 2 \] \[ z_F = 4 \times 1 = 4 \] So, \( F = (3, 2, 4) \).

Step 3: Find image point \( P = 2F - Q \).
\[ P_x = 2 \times 3 - 7 = 6 - 7 = -1 \] \[ P_y = 2 \times 2 - (-2) = 4 + 2 = 6 \] \[ P_z = 2 \times 4 - 5 = 8 - 5 = 3 \] So, \( P = (-1, 6, 3) \).

Step 4: Calculate vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \).
\[ \overrightarrow{PQ} = Q - P = (7 - (-1), -2 - 6, 5 - 3) = (8, -8, 2) \] \[ \overrightarrow{PR} = R - P = (5 - (-1), 5 - 6, 8 - 3) = (6, -1, 5) \]

Step 5: Compute cross product \( \overrightarrow{PQ} \times \overrightarrow{PR} \).
\[ \overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 8 & -8 & 2 \\ 6 & -1 & 5 \end{vmatrix} = \hat{i}((-8)(5) - 2(-1)) - \hat{j}(8 \times 5 - 2 \times 6) + \hat{k}(8 \times (-1) - (-8) \times 6) \] Calculate each component:
\[ \hat{i}(-40 + 2) - \hat{j}(40 - 12) + \hat{k}(-8 + 48) = \hat{i}(-38) - \hat{j}(28) + \hat{k}(40) \] \[ = (-38, -28, 40) \]

Step 6: Find the magnitude squared of the cross product.
\[ | \overrightarrow{PQ} \times \overrightarrow{PR} |^2 = (-38)^2 + (-28)^2 + 40^2 = 1444 + 784 + 1600 = 3828 \] Check if any simplification possible.

Step 7: Find square of area of \(\triangle PQR\).
Area of triangle: \[ \text{Area} = \frac{1}{2} \left|\overrightarrow{PQ} \times \overrightarrow{PR}\right| \] So, \[ \text{Area}^2 = \frac{1}{4} \times 3828 = 957 \]

Final answer:
\[ \boxed{957} \]