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?
On
The rabbit trying to escape a fox runs north for $8.0m$, darts northwest for $1.0m$, $\ldots$
Let's give this part a treatment of complex numbers. The situation will be similar to rotating the point $i$ by an angle $\frac{π}{4}$, and then shifting the origin to the point $-8i$. The new position of the rabbit is given by the complex number
$$ z = ie^{\frac{iπ}{4}} + 8i = -\frac{1}{\sqrt{2}} + \left(8+\frac{1}{\sqrt{2}}\right)i. $$
$\ldots$ then drops $1.0m$ down a hole into its burrow.
Let's revert back to vectors for this purpose. The situation then is similar to adding a component of $-\hat{k}$ to the vector $-\frac{1}{\sqrt{2}}\hat{i} +\left(8+\frac{1}{\sqrt{2}}\right)\hat{j}$. The final position is
$$ \vec{v} = -\frac{1}{\sqrt{2}}\hat{i} +\left(8+\frac{1}{\sqrt{2}}\right)\hat{j} -\hat{k}. $$
The magnitude of displacement is given by $|\vec{v}|$, where
$$ |\vec{v}| = \sqrt{v_x^2+v_y^2+v_z^2} \\ = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(8+\frac{1}{\sqrt{2}}\right)^2 + 1} \\ = \sqrt{66 +8\sqrt{2}}. $$
On
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.
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 $z$):
$$\begin{align}\text{Distance} &= \sqrt{\left( - \frac 1 {\sqrt 2}\right)^2 + \left(8+ \frac 1 {\sqrt 2}\right)^2 + (-1)^2}\\ &=\sqrt{\frac 1 2 + \left(64 + \frac 1 2 + \frac{16}{\sqrt 2}\right)+1}\\ &=\sqrt{66 + \frac{16}{\sqrt 2}}\\ &=\sqrt{66 + 8 \sqrt 2}\\ &\approx 8.79\mathrm m\end{align}$$
Hint: It is easier to find the horizontal displacement first. What is the long side of the triangle at ground level? You will have to use the law of cosines as the two sides are not perpendicular. Then note that the vector down the hole is perpendicular to this one, so you have another right triangle.
Your approach was fine, too. You had a $1-1-\sqrt 2$ right triangle from the northwest and down, but then assessing the angle with the north is difficult.
Another approach, I suspect the intended one, is just to find the final coordinates of the rabbit in $\Bbb R^3$. Then you have perpendicular sides and can use Pythagoras.