Oblique Coordinates

There are many ways to solve this problem. In this solution, I will be using using vectors.

Let the unit vectors along $+x$ and $+y$ directions be $\hat{i}$ and $\hat{j}$ respectively. Note that the angle between the two vectors is $60 ^{\circ}$ . We will calculate the value of the dot product $\hat{i} \cdot \hat{j}$ since it will be useful in the working ahead.

$\begin{aligned} \hat{i} \cdot \hat{j} & = \lvert \hat{i} \rvert \times |\hat{j} | \times \cos 60^{\circ} \\ & = 1 \times 1 \times \frac{1}{2} \\ & = \frac{1}{2} \end{aligned}$

The position vector of point $A$ is $\vec{OA} = \hat{i} + 6 \hat{ j}$ , and the position vector of point $B$ is $\vec{OB} = 5 \hat{i} + 2 \hat{j}$ . Using this, we can find the vector $\vec{AB}$ .

$\vec{AB} = \vec{OB} - \vec{OA} = 4 \hat{i} - 4 \hat{j}$

We want to find $|\vec{AB}|$ , the length of vector $\vec{AB}$ . We can do this by taking the dot product of $\vec{AB}$ with itself. We get $\vec{AB} \cdot \vec{AB} = |\vec{AB}|^{2}$ . Using the value of $\vec{AB}$ we found earlier, we can also write

$\begin{aligned} \vec{AB} \cdot \vec{AB} & = (4 \hat{i} - 4\hat{j}) \cdot (4 \hat{i} - 4\hat{j}) \\ & = 16 \hat{i} \cdot \hat{i} - 16 \hat{i} \cdot \hat{j} - 16 \hat{j} \cdot \hat{i} + 16 \hat{j} \cdot \hat{j} \end{aligned}$

We saw that $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = 1$ and $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{i} = \frac{1}{2}$ . Therefore

$|\vec{AB}|^{2} = 16 - 8 - 8 + 16 = 16$

Therefore $|\vec{AB}| = \sqrt{16} = 4$ $_\square$

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Bonus If the angle between the positive axes is $\omega$ , then find the distance between the points $(x_{1}, y_{1} )$ and $(x_{2}, y_{2})$

$B-A=(4,-4)$
Draw on the paper and you will see that $(0,0), (4,-4), (4,0)$ forms a equilateral triangle (because of the $60^\circ$ ).
$\therefore \vert(4,-4)\vert = 4$
Answer $=4$

For people who do not know vector algebra, there is a solution with coordinate geometry,(I am using the logic of a 13-year-old (me))

We can observe that if we have perpendicular lines instead of $60^\circ$ we can easily solve the problem. So let us try to find the perpendicular distance of only one point. Let the point be $P(x, y)$ in the oblique system. Draw the lines perpendicular to $P$ and oblique to $P$ . Let the Point where the oblique line meets x-axis be $A$ and the point of intersection of the perpendicular line and x-axis be $B$ .

Since $PA$ $||$ y-axis, $\angle PAB = 60^\circ$ Now, we can find $PB, OB$ (Perpendicular y, x) by trigonometry. $sin 60^\circ = PB/y$ $\Rightarrow$ $PB = \frac{\sqrt{3} \times y}{2}$ ) and $cos 60^\circ = AB/y$ $\Rightarrow$ $AB = \frac{y}{2}$ ) $\Rightarrow$ $OB = \frac{y}{2}\ + x$

Now we have found the formulas for the perpendiculars. Let us now come back to the question: Distance between $(1, 6) and (5, 2)$

Putting into the formulas perpendicular y-distance for $(1, 6)$ = $3\sqrt{3}$ and x- distance = $4$

Similarly for $(5, 2)\rightarrow$ y-distance = $\sqrt{3}$ and x - distance = $6$

Since all of these are only perpendicular distances, we can use the normal distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Putting the values, we get $4$ as the answer

If you have any suggestion or a better solution, please comment. If any part is confusing, comment and I will try to explain it further.