Question

Let \(a, b, c\in\mathbb{R}\) be such that \(a^2+b^2+c^2=1\). If
\(a\cos\theta=b\cos\left(\theta+\frac{2\pi}{3}\right)=c\cos\left(\theta+\frac{4\pi}{3}\right)\),
where \(\theta=\frac{\pi}{9}\), then the angle between the vectors \(a\hat{i}+b\hat{j}+c\hat{k}\) and \(b\hat{i}+c\hat{j}+a\hat{k}\) is :

• A. 0
• B. \(\frac{\pi}{9}\)
• C. \(\frac{2\pi}{3}\)
• D. \(\frac{\pi}{2}\)

Hint:

Set the common value equal to \(k\), express \(\cos\theta\) etc. as \(k/a,\ k/b,\ k/c\) and use \(\cos\theta+\cos(\theta+\frac{2\pi}{3})+\cos(\theta+\frac{4\pi}{3})=0\).

Answer 1

The Correct Option is D

Let \(a\cos\theta=b\cos(\theta+\frac{2\pi}{3})=c\cos(\theta+\frac{4\pi}{3})=k\). Since \(\theta=\frac{\pi}{9}\), \(\cos\theta\ne0\), so \(k\ne0\) and \(a,b,c\ne0\).
Then \(\cos\theta=\frac{k}{a},\ \cos(\theta+\frac{2\pi}{3})=\frac{k}{b},\ \cos(\theta+\frac{4\pi}{3})=\frac{k}{c}\).
The three cosines sum to zero: \(\cos\theta+\cos(\theta+\frac{2\pi}{3})+\cos(\theta+\frac{4\pi}{3})=0\).
So \(k\left(\frac1a+\frac1b+\frac1c\right)=0\Rightarrow ab+bc+ca=0\).
For \(\vec u=(a,b,c)\) and \(\vec v=(b,c,a)\): \(\vec u\cdot\vec v=ab+bc+ca=0\).
Both are non-zero (each has magnitude 1), so the angle is \(\frac{\pi}{2}\).
Answer: (4) \(\frac{\pi}{2}\)