If $\vec{b}$ and $\vec{c}$ are two non-collinear vectors such that $\vec{a}\cdot(\vec{b}+\vec{c})=4$ and $\vec{a}\times(\vec{b}\times\vec{c})=(x^2-2x+6)\vec{b}+(\sin y)\vec{c}$, then:
Step-by-Step Solution
Key Concept: Expand the vector triple product and equate coefficients of non-collinear vectors $\vec{b}$ and $\vec{c}$ to determine the values of $x$ and $y$.
Using the vector triple product formula: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$. Since $\vec{b}$ and $\vec{c}$ are non-collinear, we can equate coefficients with $(x^2-2x+6)\vec{b}+(\sin y)\vec{c}$. This gives: $\vec{a} \cdot \vec{c} = x^2-2x+6$ and $-(\vec{a} \cdot \vec{b}) = \sin y$. From $\vec{a} \cdot (\vec{b}+\vec{c}) = 4$, we have $\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 4$. For consistency, we need $x^2-2x+6 = (x-1)^2+5 \geq 5$ (minimum value at $x=1$), and since $\vec{a} \cdot \vec{b} = -\sin y$, we need $-\sin y \leq 1$. With $x=1$: $\vec{a} \cdot \vec{c} = 5$, so $\vec{a} \cdot \vec{b} = -1 = -\sin y$, giving $\sin y = 1$, thus $y = \frac{\pi}{2}$.
Correct Answer: 1,3