Maths Olympiad Prep

Library / /250 of 520

Algebra Difficulty 3.2 AMC 10/12 Find the answer

Given vectors a\overrightarrow {a} and b\overrightarrow {b} that satisfy a=1|\overrightarrow {a}|=1, b=2|\overrightarrow {b}|=\sqrt {2}, and a\overrightarrow {a} is perpendicular to (a\overrightarrow {a} + b\overrightarrow {b}), find the size of the angle between a\overrightarrow {a} and b\overrightarrow {b}.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Since a\overrightarrow {a} is perpendicular to (a\overrightarrow {a} + b\overrightarrow {b}), we have a(a+b)=a2+ab=0\overrightarrow {a} \cdot (\overrightarrow {a} + \overrightarrow {b}) = \overrightarrow {a}^2 + \overrightarrow {a} \cdot \overrightarrow {b} = 0.

Let the size of the angle between a\overrightarrow {a} and b\overrightarrow {b} be θ\theta. Then, from the problem, we get 1+2cosθ=01 + \sqrt{2}\cos\theta = 0.

Solving for cosθ\cos\theta, we get cosθ=22\cos\theta = -\frac{\sqrt{2}}{2}.

Given that 0θ<π0 \leq \theta < \pi, we find θ=3π4\theta = \frac{3\pi}{4}.

Therefore, the answer is 3π4\boxed{\frac{3\pi}{4}}.

This problem primarily tests the understanding of the definition of the dot product of two vectors, the property of perpendicular vectors, and finding an angle based on the value of a trigonometric function. It is a basic problem.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.