Question Type 4: Finding the time at which two objects are closest each other given some trajectory Exercises

For $\boldsymbol{P}(t) = \begin{pmatrix} 2 \\ 1 \\ 4 \end{pmatrix} + t \begin{pmatrix} 5 \\ 9 \\ 1 \end{pmatrix}$ and $\boldsymbol{Q}(t) = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + t \begin{pmatrix} 8 \\ 7 \\ 2 \end{pmatrix}$, find the time of closest approach and verify that $\boldsymbol{D}(t) = \boldsymbol{P}(t) - \boldsymbol{Q}(t)$ at this time is perpendicular to the relative velocity.

Two particles move along trajectories $\mathbf{r}_1(t) = \begin{pmatrix} 2 \\ 1 \\ 4 \end{pmatrix} + t \begin{pmatrix} 5 \\ 9 \\ 1 \end{pmatrix}$ and $\mathbf{r}_2(t) = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + t \begin{pmatrix} 8 \\ 7 \\ 2 \end{pmatrix}$, where $t$ is time.

Find the time $t$ at which the distance between the two particles is minimized.

Derive the formula for the time $t$ minimizing the distance between $P(t) = P_0 + t\mathbf{v}$ and $Q(t) = Q_0 + t\mathbf{w}$, then apply it to $P_0 = (2, 1, 4)$, $\mathbf{v} = (5, 9, 1)$ and $Q_0 = (1, 0, 1)$, $\mathbf{w} = (8, 7, 2)$.

Two particles $P$ and $Q$ move such that their position vectors at time $t$ are given by $P(t) = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} + t \begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix}$ and $Q(t) = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + t \begin{pmatrix} 0 \\ 4 \\ 3 \end{pmatrix}$.

Find the time $t$ at which the particles are at their point of minimum separation.

Two particles $P$ and $Q$ move such that their position vectors at time $t$ are given by: $\boldsymbol{r}_P(t) = \begin{pmatrix} 4 \\ 0 \\ 0 \end{pmatrix} + t \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ $\boldsymbol{r}_Q(t) = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}$ where $t \ge 0$ is the time in seconds.

Determine the time $t$ at which the particles are at their closest approach.

Find the time at which $P(t) = (1, 1, 0) + t(2, 3, 4)$ and $Q(t) = (0, 2, 3) + t(5, 1, -1)$ are closest.

Two points move such that their positions at time $t$ are given by $P(t) = (2t, 2t, 0)$ and $Q(t) = (1-t, 2+t, 0)$.

Find the time $t$ when the distance between the two points is smallest.

Determine the time $t$ and the corresponding distance $d$ when $P(t)=(0,0,1)+t(1,1,1)$ and $Q(t)=(1,2,3)+t(-1,2,0)$ are closest.

For the trajectories $P(t) = (1, 0, 0) + t(3, 4, 0)$ and $Q(t) = (0, 3, 0) + t(4, -3, 0)$, find the time of minimum separation and the minimum distance.

Determine the time $t$ at which the moving points $P(t)=(2,2,2)+t(1,-1,2)$ and $Q(t)=(3,0,1)+t(-2,3,1)$ attain their minimum separation.

Find the time $t$ when the distance between $P(t) = (1, 2, 3) + t(4, 0, 1)$ and $Q(t) = (2, 0, 5) + t(1, 3, 2)$ is minimized.

Two objects follow paths given by $\boldsymbol{P}(t) = \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} + t \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}$ and $\boldsymbol{Q}(t) = \begin{pmatrix} -2 \\ 4 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix}$.

Find the time $t$ at which the objects are closest to each other.