Let $\vec{a}$, $\vec{\beta}$ and $\vec{\gamma}$ be the unit vectors such that $\vec{a}$ and $\vec{\beta}$ are mutually perpendicular and $\vec{\gamma}$ is equally inclined to $\vec{a}$ and $\vec{\beta}$ at an angle $\theta$. If $\vec{\gamma}=x\vec{a}+y\vec{\beta}+z[\vec{a}\times\vec{\beta}]$, then:
Step-by-Step Solution
Key Concept: Equal inclination of $\vec{\gamma}$ to perpendicular unit vectors $\vec{a}$ and $\vec{\beta}$ forces $x = y$, and the unit vector normalization gives the relationship between $z^2$ and $x^2$, $y^2$.
Since $\vec{a}$ and $\vec{\beta}$ are orthonormal unit vectors, and $\vec{\gamma}$ is a unit vector equally inclined to both at angle $\theta$, we have $\vec{\gamma} \cdot \vec{a} = \cos\theta = \vec{\gamma} \cdot \vec{\beta}$, giving $x = y = \cos\theta$. Since $|\vec{\gamma}| = 1$: $x^2 + y^2 + z^2 = 1$. Substituting $x = y = \cos\theta$ yields $2\cos^2\theta + z^2 = 1$, so $z^2 = 1 - 2\cos^2\theta = 1 - 2x^2 = 1 - 2y^2$. From the constraint $x^2 + y^2 + z^2 = 1$ combined with $x = y$, all four statements follow: options 1, 2, 3 are equivalent forms of the unit vector condition, and option 4 directly from the equal inclination property.
Correct Answer: 1,2,3,4