Answer to Question #95636 in Mechanics | Relativity for Moses

Question

A man start from a point A and walks a distance of 3.0km due Northeast to point B. From point B he then walk 4.0 km due Southeast to point X.
Calculate the shortest distance between A and X.

Answer 1

Let's choose the $Ox$ axis along the east direction and $Oy$ axis along the north direction and let $\varphi$ be the angle with axis $Ox$. Then unit vector in direction with angle $\varphi$ will have coordinates ${{\vec e}_\varphi } = (\cos \varphi ,\sin \varphi )$. Northeas is the direction with ${\varphi _1} = \frac{\pi }{4}$ in our system, for Southeast we've got ${\varphi _2} = - \frac{\pi }{4}$ . Then we can find the displacement vector as

$$\vec d = {l_1}{{\vec e}_{{\varphi _1}}} + {l_2}{{\vec e}_{{\varphi _2}}}$$

where ${l_1}$ is the travelled distance along northeast direction and ${l_2}$ is the travelled distance along the southeast direction. The shortest distance is the absolute value of ${\vec d}$ . So let's find the vector

$$\vec d = 3[{\text{km}}] \cdot (\cos \frac{\pi }{4},\sin \frac{\pi }{4}) + 4[{\text{km}}](\cos ( - \frac{\pi }{4}),\sin ( - \frac{\pi }{4}))$$

Perform thecalculations

$$\vec d = 3[{\text{km}}] \cdot (\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}) + 4[{\text{km}}](\frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }})$$

$$\vec d = \frac{1}{{\sqrt 2 }}(3,3)[{\text{km}}] + \frac{1}{{\sqrt 2 }}(4, - 4)[{\text{km}}] = \frac{1}{{\sqrt 2 }}(7, - 1)[{\text{km}}]$$

And find the absolute value

$$\left| {\vec d} \right| = \frac{1}{{\sqrt 2 }}\sqrt {{7^2} + {{( - 1)}^2}} [{\text{km}}] = \frac{1}{{\sqrt 2 }}\sqrt {{7^2} + {{( - 1)}^2}} [{\text{km}}] = \frac{{5\sqrt 2 }}{{\sqrt 2 }}[{\text{km}}] = 5[{\text{km}}]$$

So the $\left| {\vec d} \right| = 5[{\text{km}}]$ is the answer.