Question

Let the image of the point \(P(0,-5,0)\) in the line \[ \frac{x-1}{2}=\frac{y}{1}=\frac{z+1}{-2} \] be the point \(R\) and the image of the point \(Q(0,-\frac12,0)\) in the line \[ \frac{x-1}{-1}=\frac{y+9}{4}=\frac{z+1}{1} \] be the point \(S\). Then the square of the area of the parallelogram \(PQRS\) is ______.

Answer 1

Correct Answer: 100

To find the images of points \(P\) and \(Q\) on their respective lines, we first determine their projections. The line for point \(P\) is given by \(\frac{x-1}{2}=\frac{y}{1}=\frac{z+1}{-2}\). We can write the parametric equations as: \[ x = 1 + 2t, \quad y = t, \quad z = -1 - 2t \] The projection of point \(P(0,-5,0)\) onto the line requires finding \(t\) such that the vector from the line to \(P\) is perpendicular to the direction vector \( \langle 2, 1, -2 \rangle \): \[ \langle 0 - (1+2t), -5 - t, 0 - (-1-2t) \rangle \cdot \langle 2, 1, -2 \rangle = 0 \] \[ \langle -1-2t, -5-t, 1+2t \rangle \cdot \langle 2, 1, -2 \rangle = 0 \] \[ -2 - 4t - 5 - t - 2 - 4t = 0 \] \[ -12 - 9t = 0 \Rightarrow t = -\frac{4}{3} \] Substitute back to find the projection coordinates: \[ x = 1 - \frac{8}{3} = -\frac{5}{3}, \quad y = -\frac{4}{3}, \quad z = \frac{5}{3} \] The image \(R\) is found by reflecting \(P\) across the line; hence, the coordinates double the difference: \[ R = \left( 2\left(-\frac{5}{3}\right), 2\left(-\frac{4}{3}\right)+1, 2\left(\frac{5}{3}\right)-1 \right) \] \[ R = \left(-\frac{10}{3}, -\frac{8}{3}, \frac{7}{3}\right) \] Similarly, for point \(Q(0,-\frac{1}{2},0)\) in the line \(\frac{x-1}{-1}=\frac{y+9}{4}=\frac{z+1}{1}\): \[ x = 1-t, \quad y = -9+4t, \quad z = -1+t \] Set up the projection: \[ \langle 0-(1-t), -\frac{1}{2}-(-9+4t), 0-(-1+t) \rangle \cdot \langle -1, 4, 1 \rangle = 0 \] \[ \langle -1+t, \frac{17}{2}-4t, -1+t \rangle \cdot \langle -1, 4, 1 \rangle = 0 \] \[ (1-t) + 2(\frac{17}{2}-4t) -1+t = 0 \] \[ 1-t+17-8t-1+t = 0 \] \[ 17-8t = 0 \Rightarrow t = \frac{17}{8} \] Calculate the projection: \[ x = 1-\frac{17}{8} = -\frac{9}{8}, \quad y = -9+4\left(\frac{17}{8}\right) = -\frac{25}{2}, \quad z = -1+\frac{17}{8} = \frac{9}{8} \] Reflect to find \(S\): \[ S = \left(-\frac{9}{4}, -\frac{25}{2}, \frac{9}{4}\right) \] Now, compute the area of parallelogram \(PQRS\). First, vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{PS}\): \[ \overrightarrow{PQ} = \left(0-0, -\frac{1}{2}+5, 0-0\right) = \langle 0, \frac{9}{2}, 0 \rangle \] \[ \overrightarrow{PS} = \left(-\frac{9}{4} - 0, -\frac{25}{2} - 5, \frac{9}{4} - 0\right) \] \[ \overrightarrow{PS} = \left(-\frac{9}{4}, -\frac{35}{2}, \frac{9}{4}\right) \] Compute the cross product \(\overrightarrow{PQ} \times \overrightarrow{PS}\): \[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & \frac{9}{2} & 0 \\ -\frac{9}{4} & -\frac{35}{2} & \frac{9}{4} \end{vmatrix} \] \[ = \mathbf{i} \left(\frac{9}{4}\cdot\frac{9}{4}\right) - \mathbf{j} (0) + \mathbf{k}\left(-\frac{9}{4}\cdot\frac{9}{2}\right) \] \[ = \mathbf{i}\frac{81}{16} + \mathbf{k}\left(-\frac{81}{8}\right) \] The magnitude gives the area of the parallelogram: \[ = \sqrt{\left(\frac{81}{16}\right)^2 + 0^2 + \left(-\frac{81}{8}\right)^2 } = \frac{81}{8}\sqrt{\frac{5}{4}} = \frac{81\sqrt{5}}{16} \] Square of the area: \[ \left( \frac{81\sqrt{5}}{16} \right)^2 = \frac{6561\cdot 5}{256} = \frac{32805}{256} \] Therefore, the square of the area of parallelogram \(PQRS\) is \(\boxed{\frac{32805}{256}}\), satisfying the provided range.