vectors question (1 Viewer)

elmio023 · Oct 15, 2023

Can someone give the answer for ii? there were no answers attached

jimmysmith560 · Oct 15, 2023

Would the following working help?

Luukas.2 · Oct 15, 2023

$2\cos{\alpha} - (1 - 2k) = 0$ rearranges to give $\cos{\alpha} = \frac{1 - 2k}{2}$ .

Since we know that $k\in\[0,\ 1]$ , it follows that $\cos{\alpha}\in\left[-\frac{1}{2},\ \frac{1}{2}\right] \quad \implies \quad \alpha \in \left[\frac{\pi}{3},\ \frac{2\pi}{3}\right]$ .

The original form of this questions asks the values for there to be two distinct positions for M.

The equation given is always satisfied by $k = \frac{1}{2}$ .

So, for a given $\alpha$ , there are two positions for M, the one where $k = \frac{1}{2}$ and the one where $k = \frac{1 - 2\cos{\alpha}}{2}$

...except when these are both the same, which occurs when $\alpha = \frac{\pi}{2}$ . Thus, there are two possible positions for M when

$\alpha \in \left[\frac{\pi}{3},\ \frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\ \frac{2\pi}{3}\right]$

Make a Difference – Donate to Bored of Studies!

vectors question (1 Viewer)

elmio023

New Member

jimmysmith560

Le Phénix Trilingue

Luukas.2

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)