Let $\vec{a},\vec{b},\vec{c}$ be three vectors such that $\vec{a}\times\vec{b}=2(\vec{a}\times\vec{c})$. If $|\vec{a}|=1$, $|\vec{b}|=4$, $|\vec{c}|=2$, and the angle between $\vec{b}$ and $\vec{c}$ is $60^\circ$, then $|\vec{a}\cdot\vec{c}|$ is equal to:
Let A, B, C be three points in xy-plane, whose position vector are given by \sqrt3^i + ^j, ^i + \sqrt3^j and a^i + (1 - a)^j respectively with respect to the origin O . If the distance of the point C from the line bisecting the angle between - -\to - -\to the vectors OA and OB is , then the sum of all the possible values of a is : 9 \sqrt2
If the position vectors of the vertices A, B and C of a \(\triangle ABC\) are, respectively, \(4\hat{i}+7\hat{j}+8\hat{k}\), \(2\hat{i}+3\hat{j}+4\hat{k}\) and \(2\hat{i}+5\hat{j}+7\hat{k}\), then the position vector of the point, where the bisector of \(\angle A\) meets BC is
Let \(\vec{a}, \vec{b}, \vec{c}\) be three vectors of magnitude 2, 3, 5 respectively, satisfying \(|[\vec{a},\, \vec{b},\, \vec{c}]| = 30\). If \((2\vec{a} + \vec{b} + \vec{c}) \cdot ((\vec{a} \times \vec{c}) \times (\vec{a} - \vec{c}) + \vec{b}) = k\), then the value of \(\left\lfloor \dfrac{k}{103} \right\rfloor\) is: