The initial and final velocity vectors of a body are vec v_0…

The initial and final velocity vectors of a body are $\vec v_0 = A\vec i + B\vec j + C\vec k$ and $\vec v_h = E\vec i + F\vec j + G\vec k$, with $A = 2$, $B = 5$, $C = 3$, $E = 1$, $F = 5$, $G = 7$ (all in m/s). (a) Draw the velocity vectors in a Cartesian coordinate system. (b) Find the velocity change vector. (c) Find its magnitude.

Explanation of Method

Change of a velocity vector in 3D.

Step 1: (a) The vectors

Drawn from the origin, $\vec v_0$ ends at the point $(2, 5, 3)$ and $\vec v_h$ at $(1, 5, 7)$.

$x$ $y$ $z$ $(2,\,5,\,3)$ $(1,\,5,\,7)$ $\vec v_0$ $\vec v_h$ $\Delta\vec v = -\vec i + 4\vec k$ $O$

initial velocity
final velocity
velocity change

Both vectors start at the origin (units m/s). The change $\Delta\vec v$ is the arrow from the tip of $\vec v_0$ to the tip of $\vec v_h$, so that $\vec v_0 + \Delta\vec v = \vec v_h$. Dashed lines drop the tips to the $xy$-plane to help read the 3D picture.

Step 2: (b) Velocity change

Subtract component by component: $\Delta\vec v = \vec v_h - \vec v_0 = (1 - 2)\,\vec i + (5 - 5)\,\vec j + (7 - 3)\,\vec k$.

\[\Delta\vec v = -\vec i + 4\,\vec k\ \ (\mathrm{m/s})\]

Step 3: (c) Magnitude

\[ |\Delta\vec v| = \sqrt{(-1)^2 + 0^2 + 4^2} = \sqrt{17}. \]

\[|\Delta\vec v| = \sqrt{17}\ \mathrm{m/s} \approx 4.12\ \mathrm{m/s}\]

Check

$ \vec v_0 + \Delta\vec v = (2 - 1)\,\vec i + 5\,\vec j + (3 + 4)\,\vec k = \vec v_h\ \checkmark $

Result

\[\boxed{\,\Delta\vec v = -\vec i + 4\,\vec k\ \ (\mathrm{m/s})\,}\]

\[\boxed{\,|\Delta\vec v| = \sqrt{17}\ \mathrm{m/s} \approx 4.12\ \mathrm{m/s}\,}\]

Remark

Note. The magnitude of the change is not the change of the magnitudes: $|\vec v_0| = \sqrt{38} \approx 6.16$ and $|\vec v_h| = \sqrt{75} \approx 8.66\ \mathrm{m/s}$ differ by only 2.5 m/s, while $|\Delta\vec v| \approx 4.12\ \mathrm{m/s}$, because the direction changed too.