Calculating the distance between two points represented by position vectors

Question

Q: The positions of two particles, A and B, at a time t, are given by rA = 3 cos(πt)i +(2t+1)j and rB = 2sin(πt)i + (2-t)j

Find the distance between the two particles when t = 4:

My method:

When t = 4: rA = 3i + 9j and rB = -2j

the distance from the origin to point rA when t = 4 is 3 root (10)

and the distance from the origin to point rB when t = 4 is 2

if we let theta be the angle that point rA (when t = 4) is from the x-axis, then theta = arctan(3) = 71.565 degrees

We can draw a triangle with the vertices being rA, rB and the origin respectively.

Using the Cosine rule: a^2 = b^2 + c^2 - 2bccosA a^2 = (3root(10))^2 + 2^2 - 12root(10)cos(71.565)

a^2 = 82

a = root (82) metres

But the answer is apparently root (130) metres. Have I gone wrong (and if so where have I) or is the answer given incorrect?

Answer 1

The simplest method is to use $$ || \textbf{AB} || = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} = \sqrt{(-3)^2 + (-11)^2} = \sqrt{130} \approx 11,40 $$ Take a look here.

Answer 2

You used the wrong angle: you need the angle between the two vectors, not the angle between one of them and the $x$-axis. Computing this angle explicitly is unnecessary. For the law of cosines, you need $\lVert\vec r_a\rVert \lVert\vec r_b\rVert \cos\theta$, but that’s exactly $\vec r_a\cdot\vec r_b$. So, $$\lVert \vec r_a-\vec r_b\rVert^2 = \lVert\vec r_a\rVert^2+\lVert\vec r_b\rVert^2-2(\vec r_a\cdot\vec r_b),$$ which, if you plug in the values that you’ve got, agrees with the given solution.

As another answer points out, though, this is also more work than necessary: you should just compute the norm of $\vec r_a-\vec r_b$ directly.