Question (1 Viewer)

carrotsss · Oct 15, 2023

I had a stroke reading that what

scaryshark09 · Oct 15, 2023

SB257426 · Oct 15, 2023

another killara paper?
damn......

=)(= · Oct 15, 2023

SB257426 said:
another killara paper?
damn......

yea lol 2023

=)(= · Oct 15, 2023

scaryshark09 said:
A

How tho?

member 6003 · Oct 15, 2023

bro is this question even correct? like the wording makes no sense to me

SB257426 · Oct 15, 2023

=)(= said:
yea lol 2023

lol that shits difficult ngl

Luukas.2 · Oct 15, 2023

The answer is (D)

Luukas.2 · Oct 15, 2023

Take a and b as unit vectors from the origin (O), representing points A and B on the unit circle. The angle between the vectors is $2\theta$ . Since the distance AB is the length of the vector a - b and it is less than one, the isosceles triangle AOB must have the radii as its longest sides, meaning the angle $2\theta$ is no more than 60 degrees. Hence, the solution should be:

$0 \le 2\theta < \frac{\pi}{3} \qquad \implies \qquad \theta \in \left[0,\ \frac{\pi}{6}\right)$

Unfortunately, this isn't provided as an option. However, since $2\theta \in [0,\ 2\pi]$ , the length of vector a - b shortens again as $2\theta$ exceeds pi, and so there is another set of solutions for

$\frac{5\pi}{3} < 2\theta \le 2\pi \qquad \implies \qquad \theta \in \left(\frac{5\pi}{6},\ \pi\right]$

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