Solveeit Logo
Login

Question

Question: If $\overline{a}$ and $\overline{b}$ are two unit vectors such that $5\overline{a}+4\overline{b}$ an...

If a‾\overline{a} and b‾\overline{b} are two unit vectors such that 5a‾+4b‾5\overline{a}+4\overline{b} and a‾−2b‾\overline{a}-2\overline{b} are perpendicular to each other, then the angle between a‾\overline{a} and b‾\overline{b} is

A

π3\frac{\pi}{3}

B

cos−1(23)cos^{-1}(\frac{2}{3})

C

2π3\frac{2\pi}{3}

D

cos−1(13)cos^{-1}(\frac{1}{3})

Answer

2π3\frac{2\pi}{3}

Explanation

Solution

Given that (5a+4b)⋅(a−2b)=0,(5\mathbf{a}+4\mathbf{b})\cdot (\mathbf{a}-2\mathbf{b})=0, expanding the dot product using a⋅a=1, b⋅b=1,\mathbf{a}\cdot\mathbf{a}=1,\ \mathbf{b}\cdot\mathbf{b}=1, and a⋅b=cos⁡θ\mathbf{a}\cdot\mathbf{b}=\cos\theta: 5(a⋅a)−10(a⋅b)+4(b⋅a)−8(b⋅b)=5−10cos⁡θ+4cos⁡θ−8=0.5(\mathbf{a}\cdot\mathbf{a}) -10(\mathbf{a}\cdot\mathbf{b}) + 4(\mathbf{b}\cdot\mathbf{a}) -8(\mathbf{b}\cdot\mathbf{b}) = 5 -10\cos\theta + 4\cos\theta -8=0. This simplifies to: −3−6cos⁡θ=0⟹cos⁡θ=−12.-3 -6\cos\theta = 0 \quad \Longrightarrow \quad \cos\theta = -\frac{1}{2}. Thus, θ=cos⁡−1(−12)=2π3.\theta = \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}.