To solve the problem, we need to determine the time \( t \) at which the bird flying from point \( P(4m, -1m, 5m) \) sees points \( P_1(-1m, -1m, 0m) \) and \( P_2(3m, -1m, -3m) \) as collinear while flying perpendicular to the line segment \( P_1P_2 \) at a constant speed of \( 2m/s \).
### Step-by-Step Solution:
1. **Determine the Vector from \( P_1 \) to \( P_2 \)**:
\[
\vec{P_1P_2} = P_2 - P_1 = (3 - (-1), -1 - (-1), -3 - 0) = (4, 0, -3)
\]
2. **Find the Direction Vector Perpendicular to \( \vec{P_1P_2} \)**:
A vector perpendicular to \( \vec{P_1P_2} \) can be obtained by taking the cross product of \( \vec{P_1P} \) and \( \vec{P_1P_2} \). First, we find \( \vec{P_1P} \):
\[
\vec{P_1P} = P - P_1 = (4 - (-1), -1 - (-1), 5 - 0) = (5, 0, 5)
\]
3. **Calculate the Cross Product**:
\[
\vec{v} = \vec{P_1P} \times \vec{P_1P_2} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
5 & 0 & 5 \\
4 & 0 & -3
\end{vmatrix}
\]
\[
= \hat{i}(0 \cdot (-3) - 5 \cdot 0) - \hat{j}(5 \cdot (-3) - 5 \cdot 4) + \hat{k}(5 \cdot 0 - 0 \cdot 4)
\]
\[
= 0\hat{i} - (-15 - 20)\hat{j} + 0\hat{k} = 35\hat{j}
\]
Thus, the perpendicular direction vector is \( \vec{v} = (0, 35, 0) \).
4. **Normalize the Direction Vector**:
The magnitude of \( \vec{v} \) is:
\[
|\vec{v}| = \sqrt{0^2 + 35^2 + 0^2} = 35
\]
The unit vector in the direction of \( \vec{v} \) is:
\[
\hat{v} = \left(0, \frac{35}{35}, 0\right) = (0, 1, 0)
\]
5. **Calculate the Position of the Bird at Time \( t \)**:
The position of the bird at time \( t \) is given by:
\[
P(t) = P + t \cdot \text{speed} \cdot \hat{v} = (4, -1, 5) + t \cdot 2 \cdot (0, 1, 0) = (4, -1 + 2t, 5)
\]
6. **Set Up the Collinearity Condition**:
For points \( P_1, P_2, \) and \( P(t) \) to be collinear, the vectors \( \vec{P_1P(t)} \) and \( \vec{P_1P_2} \) must be parallel. This means:
\[
\vec{P_1P(t)} = P(t) - P_1 = (4 - (-1), -1 + 2t - (-1), 5 - 0) = (5, 2t, 5)
\]
The condition for collinearity is:
\[
\frac{5}{4} = \frac{2t}{0} \quad \text{(This is not valid, so we need to check the other dimension)}
\]
7. **Using the Z-Coordinate**:
Since \( P_1 \) and \( P_2 \) have the same y-coordinate, we can ignore the y-component. We check the z-component:
\[
\text{For collinearity, we need: } \frac{5}{-3} = \frac{5}{0} \quad \text{(This is also invalid)}
\]
8. **Finding Time \( t \)**:
Since the bird is flying perpendicular to \( P_1P_2 \), we can find the distance it needs to travel to reach the line formed by \( P_1 \) and \( P_2 \). The distance from \( P \) to the line can be calculated using the formula for the distance from a point to a line in 3D.
After calculating the distance and dividing by the speed, we find:
\[
t = \frac{\text{Distance}}{\text{Speed}} = \frac{7}{2} = 3.5 \text{ seconds}
\]
### Final Answer:
The time \( t \) at which the bird sees \( P_1 \) and \( P_2 \) as collinear is \( 3.5 \) seconds.
