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Let $$\vec{a}$$ and $$\vec{b}$$ be two unit vectors such that $$\vec{a}\cdot \vec{b}=0$$. For some x, y $$\epsilon\ \mathbb{R}$$, let $$\vec{c}=x\vec{a}+y\vec{b}+(\vec{a}\times \vec{b})$$. If $$|\vec{c}|=2$$ and the vector $$\vec{c}$$ is inclined at the same angle $$\alpha$$ to both $$\vec{a}$$ and $$\vec{b}$$, then the value of $$8\cos^2\alpha$$ is __________.


Solution

$$\vec{c}=x\vec{a}+y\vec{b}+\vec{a}\times \vec{b}$$
$$\vec{c}\cdot \vec{a}=x$$ and $$x=2\cos\alpha$$
$$\vec{c}\cdot \vec{b}=y$$ and $$y=2\cos\alpha$$
Also, $$|\vec{a}\times \vec{b}|=1$$

$$\therefore \vec{c}=(2\cos\alpha )(\vec{a}+\vec{b})+\vec{a}\times \vec{b}$$

$$\therefore |\vec{c}|^2=4\cos^2\alpha(\vec{a}+\vec{b})^2+(\vec{a}\times \vec{b})^2+2\cos\alpha (\vec{a}+\vec{b})\cdot (\vec{a}\times \vec{b})$$

$$\therefore 4=8\cos^2\alpha +1$$

$$\therefore 8\cos^2\alpha =3$$

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