To solve this problem, we need to determine the shortest distance $d$ from the point $P_0 = (-4, 4, 3)$ to the line $L$ defined by the point $P = (5, -1, -1)$ and the direction vector $\vec{d} = [0, 1, -2]^T$. Additionally, we need to find the point $Q$ on $L$ that is closest to $P_0$.

Solution Steps

1. Parametrize the line $L$: The line $L$ passing through $P = (5, -1, -1)$ with direction vector $\vec{d} = [0, 1, -2]^T$ can be written as: $L(t) = P + t \vec{d} = (5, -1, -1) + t(0, 1, -2)$ So, the parametric form of $L$ is: $L(t) = (5, -1 + t, -1 - 2t)$

2. Vector from $P_0$ to $P$: Let $\vec{P_0P}$ be the vector from $P_0$ to $P$: $\vec{P_0P} = P - P_0 = (5 - (-4), -1 - 4, -1 - 3) = (9, -5, -4)$

3. Project $\vec{P_0P}$ onto $\vec{d}$: The projection of $\vec{P_0P}$ onto $\vec{d}$ is given by: $\text{proj}_{\vec{d}} \vec{P_0P} = \frac{\vec{P_0P} \cdot \vec{d}}{\vec{d} \cdot \vec{d}} \vec{d}$

   • Calculate $\vec{P_0P} \cdot \vec{d}$: $\vec{P_0P} \cdot \vec{d} = (9)(0) + (-5)(1) + (-4)(-2) = 0 - 5 + 8 = 3$
   • Calculate $\vec{d} \cdot \vec{d}$: $\vec{d} \cdot \vec{d} = (0)^2 + (1)^2 + (-2)^2 = 0 + 1 + 4 = 5$
   • Therefore, $\text{proj}_{\vec{d}} \vec{P_0P} = \frac{3}{5} \vec{d} = \frac{3}{5}(0, 1, -2) = (0, \frac{3}{5}, -\frac{6}{5})$

4. Calculate $\vec{P_0Q}$: The vector from $P_0$ to the closest point $Q$ on $L$ is the component of $\vec{P_0P}$ orthogonal to $\vec{d}$: $\vec{P_0Q} = \vec{P_0P} - \text{proj}_{\vec{d}} \vec{P_0P} = (9, -5, -4) - (0, \frac{3}{5}, -\frac{6}{5})$ $\vec{P_0Q} = \left(9, -5 - \frac{3}{5}, -4 + \frac{6}{5}\right) = \left(9, -\frac{28}{5}, -\frac{14}{5}\right)$

5. Distance $d$: The shortest distance $d$ from $P_0$ to $L$ is the magnitude of $\vec{P_0Q}$: $d = \left|\vec{P_0Q}\right| = \sqrt{9^2 + \left(-\frac{28}{5}\right)^2 + \left(-\frac{14}{5}\right)^2}$ $= \sqrt{81 + \frac{784}{25} + \frac{196}{25}} = \sqrt{\frac{2025}{25}} = \sqrt{120}$

6. Point $Q$ on $L$: To find $Q$, substitute $t = \frac{3}{5}$ (the value found from the projection calculation) into the parametric equation of $L$: $Q = (5, -1 + \frac{3}{5}, -1 - 2 \cdot \frac{3}{5}) = \left(5, -\frac{2}{5}, -\frac{11}{5}\right)$

Final Answer

$d = \sqrt{120}$ $Q = \left(5, -\frac{2}{5}, -\frac{11}{5}\right)$

Would you like additional details on these steps?

1. What is the formula for projecting one vector onto another?
2. How can you find the shortest distance from a point to a line in three dimensions?
3. What does it mean for two vectors to be orthogonal?
4. How is the parametric equation of a line derived from a point and direction vector?
5. Why is the distance to the line equal to the magnitude of the orthogonal vector component?

Tip: For finding the shortest distance from a point to a line, use vector projection to find the closest point on the line and then compute the orthogonal distance directly.