Two insects A and B fly in space with uniform velocities VA = i + 4j + 3k and VB = 4i + 2j − 4k in m/s with respect to a stationary observer at the origin (0,0,0). Show that the insects fly at right angle to each other and determine their distance apart after 5s. (39.37 m).

The velocity of insect A is $\vec{V_A} = \vec{i} + 4\vec{j} + 3\vec{k}$ m/s.
The velocity of insect B is $\vec{V_B} = 4\vec{i} + 2\vec{j} - 4\vec{k}$ m/s.

If two vectors are perpendicular, their dot product is zero.

Since the dot product of $\vec{V_A}$ and $\vec{V_B}$ is zero, the velocity vectors are perpendicular. Therefore, the insects fly at a right angle to each other.

The relative velocity of insect B with respect to insect A is $\vec{V_{BA}} = \vec{V_B} - \vec{V_A}$.

Assuming both insects start at the origin $(0,0,0)$ at $t=0$, their relative displacement after time $t$ is $\vec{r_{BA}}(t) = \vec{V_{BA}} t$.
For $t = 5$ s:

The distance apart is the magnitude of $\vec{r_{BA}}(5)$.

The distance apart after 5 seconds is: