To solve the problem, let's analyze the given information and apply relevant circle geometry principles.
The arc AC subtends a right angle, making it a quarter of the circle. We know:
Let the length of arc AB be \(x\) and the length of arc BC be \(y\). Thus, \(x/y = 1/5\), giving \(y = 5x\).
Since the total arc AC subtends a right angle (90 degrees),
Substituting \(y = 5x\) into the equation:
Thus, the angle subtended by arc AB at the center \(O\) is \(15^\circ\) and by arc BC is \(75^\circ\).
Now, use vector analysis: \( \overrightarrow{OC} = \alpha \overrightarrow{OA} + \beta \overrightarrow{OB} \). Since points \(A\), \(B\), and \(C\) are on the circle, we can convert the angular measure into the unit circle coordinates:
Substitute these into the equation:
Solve the equations:
The expected form \(\alpha = \sqrt{2} (\sqrt{3}-1) \beta\) gives:
Thus, the correct answer is \(\boxed{2-\sqrt{3}}\).
If \( \vec{u}, \vec{v}, \vec{w} \) are non-coplanar vectors and \( p, q \) are real numbers, then the equality:
\[ [3\vec{u} \quad p\vec{v} \quad p\vec{w}] - [p\vec{v} \quad \vec{w} \quad q\vec{u}] - [2\vec{w} \quad q\vec{v} \quad q\vec{u}] = 0 \]
holds for:
Let \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) be position vectors of three non-collinear points on a plane. If
\[ \alpha = \left[\mathbf{a} \quad \mathbf{b} \quad \mathbf{c}\right] \text{ and } \mathbf{r} = \mathbf{a} \times \mathbf{b} - \mathbf{c} \times \mathbf{b} - \mathbf{a} \times \mathbf{c}, \]
Then \(\frac{|\alpha|}{|\mathbf{r}|}\) represents:
If
\[ P = (a \times \mathbf{i})^2 + (a \times \mathbf{j})^2 + (a \times \mathbf{k})^2 \]
and
\[ Q = (a \cdot \mathbf{i})^2 + (a \cdot \mathbf{j})^2 + (a \cdot \mathbf{k})^2, \]
Then find the relation between \(P\) and \(Q\).
Given vectors \(\mathbf{a} = \mathbf{i} + \mathbf{j} - 2\mathbf{k}\), \(\mathbf{b} = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k}\), \(\mathbf{c} = 2\mathbf{i} - \mathbf{j} + \mathbf{k}\), and \(\mathbf{r}\) such that
\[ \mathbf{r} \cdot \mathbf{a} = 0, \\ \mathbf{r} \cdot \mathbf{c} = 3, \\ [\mathbf{r} \quad \mathbf{a} \quad \mathbf{b}] = 0, \]
Then find \(|\mathbf{r}|\).