A mosquito net positioned over a 7 ft × 4 ft bed has a height of 3...

The origin is taken at the hole where the mosquito enters, so the initial position vector is $\vec{r}_1 = 0\hat{i} + 0\hat{j} + 0\hat{k}$.

The mosquito flies to the diagonally opposite upper corner of the net.

The length of the bed is along the X-axis, so the x-component of the displacement is $7 \text{ ft}$.

The width of the bed is along the Y-axis, so the y-component of the displacement is $4 \text{ ft}$.

The vertically upward direction is along the Z-axis, so the z-component of the displacement is $3 \text{ ft}$.

The final position vector of the mosquito is $\vec{r}_2 = 7\hat{i} + 4\hat{j} + 3\hat{k}$.

The displacement vector is $\Delta\vec{r} = \vec{r}_2 - \vec{r}_1 = 7\hat{i} + 4\hat{j} + 3\hat{k}$.

The components of the displacement vector are $7 \text{ ft}, 4 \text{ ft}, 3 \text{ ft}$.