The equation of a circle which passes through the points of intersection of the circles \[ 2x^2 + 2y^2 - 2x + 6y - 3 = 0, \quad x^2 + y^2 + 4x + 2y + 1 = 0 \] and whose centre lies on the common chord of these circles is:
The equation of the circle passing through the origin and cutting the circles $x^2 + y^2 + 6x - 15 = 0$ and $x^2 + y^2 - 8y - 10 = 0$ orthogonally is:
S = (-1,1) is the focus, \( 2x - 3y + 1 = 0 \) is the directrix corresponding to S and \( \frac{1}{2} \) is the eccentricity of an ellipse. If \( (a,b) \) is the centre of the ellipse, then \( 3a + 2b \) is:
Let \( \bar{a} \) be a vector perpendicular to the plane containing non-zero vectors \( \bar{b} \) and \( \bar{c} \). If \( \bar{a}, \bar{b}, \bar{c} \) are such that
\[ |\bar{a} + \bar{b} + \bar{c}| = \sqrt{|\bar{a}|^2 + |\bar{b}|^2 + |\bar{c}|^2}, \]
then
\[ |(\bar{a} \times \bar{b}) \cdot (\bar{a} \times \bar{c})| = \]
Let \( \bar{a}, \bar{b}, \bar{c} \) be three vectors each having \( \sqrt{2} \) magnitude such that
\[ (\bar{a}, \bar{b}) = (\bar{b}, \bar{c}) = (\bar{c}, \bar{a}) = \frac{\pi}{3}. \]
If
\[ \bar{x} = \bar{a} \times (\bar{b} \times \bar{c}) \quad \text{and} \quad \bar{y} = \bar{b} \times (\bar{c} \times \bar{a}), \]
then:
\( \vec{a}, \vec{b}, \vec{c} \) are three vectors such that \(|\vec{a}| = 3\), \(|\vec{b}| = 2\sqrt{2}\), \(|\vec{c}| = 5\), and \( \vec{c} \) is perpendicular to the plane of \( \vec{a} \) and \( \vec{b} \).
If the angle between the vectors \( \vec{a} \) and \( \vec{b} \) is \( \frac{\pi}{4} \), then
\[ |\vec{a} + \vec{b} + \vec{c}| = \ ? \]
Given the vectors:
\[ \mathbf{a} = \mathbf{i} + 2\mathbf{j} + \mathbf{k} \]
\[ \mathbf{b} = 3(\mathbf{i} - \mathbf{j} + \mathbf{k}) = 3\mathbf{i} - 3\mathbf{j} + 3\mathbf{k} \]
where
\[ \mathbf{a} \times \mathbf{c} = \mathbf{b} \]
\[ \mathbf{a} \cdot \mathbf{x} = 3 \]
Find:
\[ \mathbf{a} \cdot (\mathbf{x} \times \mathbf{b} - \mathbf{c}) \]