\(A,B,C,D\) are four points in a plane with position vectors \(\vec{a},\vec{b},\vec{c},\vec{d}\) respectively such that \((\vec{a}-\vec{d})\cdot(\vec{b}-\vec{c})=0\) and \((\vec{b}-\vec{d})\cdot(\vec{c}-\vec{a})=0\). Then \(D\) is the
For \(p>0\), the vector \(\vec{v}_2=(2,\,-(\sqrt{3}\,p+1))\) is obtained by rotating \(\vec{v}_1=\sqrt{3}(-1,\,-(p^2+\sqrt{3}))\) about the origin counter-clockwise. If the angle of rotation is \(\theta\), find \(\tan\theta\).
Let a vector \(\hat{i}+\sqrt{2}\,\hat{j}+\sqrt{2}\,\hat{k}\) be obtained by rotating the vector \(\sqrt{3}\,\hat{j}\) by an angle \(45°\) about the origin in the clockwise direction to the first quadrant. Then the area of the triangle formed by the vector \((\hat{i}+\sqrt{2}\,\hat{j}+\sqrt{2}\,\hat{k})\) with the coordinate axes is equal to:
Consider points \(A,B,C\) with position vectors \(\vec{a},\vec{b},\vec{c}\) respectively. Statement-1: \(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\vec{0}\). Statement-2: \(A,B,C\) form the vertices of a triangle.
\(\vec{a}, \vec{b}, \vec{c}\) are 3 vectors, such that \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\), \(|\vec{a}| = 1\), \(|\vec{b}| = 2\), \(|\vec{c}| = 3\), then \(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}\) is equal to