Angles between equally nullifying vectors.

Question

Q. The vectors P,Q and R are such that |P|=|Q|, |R|=$\sqrt{2}$|P| and P+Q+R=$0$. The angles between P and Q,Q and R, P and R are respectively:

a)90,135,135

b)90,45,45

c)45,90,90

d)45,135,135

My text marks the answer as 'a'. But isn't 'b' right too. In case of 'a' we have three vectors pointing radially outward from a common origin, whereas in the case of 'b' we have three vectors forming an isosceles, right triangle with 'R' as the hypotenuse with head and tail connection forming a equilibrium vector triangle.

Am I wrong somewhere?

Answer 1

The angle between -Q and R is 45 degrees, and the angle between P and -R is 45 degrees.

If you shift the vectors so they have the same initial point, you will see that the angle between P and R is 135 degrees, as is the angle between Q and R.

Answer 2

Let's assume that b is correct, and that $P$ is not the zero vector. We know that $P + Q + R = 0$, so let's dot both sides with $P$. We get $P \cdot P + Q \cdot P + R \cdot P = 0 \cdot P = 0$. Since the angle between $P$ and $Q$ is 90 degrees, the dot product is zero. $P \cdot P$ is definitely positive. Lastly, $P \cdot R$ is positive. To see the last claim, $\theta_{P,R} = 45 = \arccos (\frac{P \cdot R}{|P||R|})$. From there you can easily show that $P \cdot R$ is positive. Thus you'd have a positive number equaling zero, which isn't possible.