[Math] Displacement vectors

physics

A rabbit trying to escape a fox runs north for 8.0m, darts northwest for 1.0m, then drops 1.0m down a hole into its burrow. What is the magnitude of the net displacement of the rabbit?

So I drew two triangles and tried to solve it using pythagorean theorem but I got stuck. I had a small triangle with two sides that are 1 and one unknown side (not sure how to find this side).

Help please?

Best Answer

You may represent the displacement as a vector $\left(\begin{matrix}x\\y\\z\end{matrix}\right)$. $x$ is the distance to the East, $y$ the distance to the North, and $z$ the distance up into the sky. I'll write it inline like $(x,y,z)$ because it's more compact.

  1. Initially, the rabbit and fox are in the same place. The displacement is $(0,0,0).$
  2. The rabbit runs north for $8 \mathrm m$. That's $8$ units in the $y$ direction. New displacement is $(0,8,0).$
  3. The rabbit darts North-West $1\mathrm m$. That's $(-1/ \sqrt 2, 1 / \sqrt 2, 0)$ (use pytharogas to work out East and North displacements of this move separately). New displacement is $(0,8,0) + (-1/\sqrt 2, 1/\sqrt 2,0) = (-1/\sqrt 2, 8 + 1/\sqrt 2, 0)$.
  4. The rabbit burrows $1 \mathrm m$ down, decreasing its $z$ coordinate by $1$. Final displacement is $$(-1/\sqrt 2, \; \; 8 + 1/\sqrt 2, \; \; -1).$$

Fortunately, Pythagorean's theorem generalises to any dimension. So to find the distance from the rabbit to the fox, we take the root of the sum of the squares of the displacements in each orthogonal direction ($x,y$ and $z$):

$$\begin{align}\text{Distance} &= \sqrt{\left( - \frac 1 {\sqrt 2}\right)^2 + \left(8+ \frac 1 {\sqrt 2}\right)^2 + (-1)^2}\\ &=\sqrt{\frac 1 2 + \left(64 + \frac 1 2 + \frac{16}{\sqrt 2}\right)+1}\\ &=\sqrt{66 + \frac{16}{\sqrt 2}}\\ &=\sqrt{66 + 8 \sqrt 2}\\ &\approx 8.79\mathrm m\end{align}$$

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