The sum of the three vectors determined by the medians of a triangle directed from the vertices is

If $|a| = 3, |b| = 4$ and $|a + b| = 5,$ then $|a - b| = $

Let $\bar{a}, \bar{b}$ and $\bar{c}$ be non-coplanar vectors. If $P, Q, R$ and $S$ are four points with position vectors $-\bar{a}+4\bar{b}-3\bar{c}$,$3\bar{a}+2\bar{b}-5\bar{c}$,$-3\bar{a}+8\bar{b}-5\bar{c}$ and $-3\bar{a}+2\bar{b}+\bar{c}$ respectively,then the ordered pair $(x, y)$ of real numbers such that $\overline{PQ} = x \cdot \overline{PR} + y \cdot \overline{PS}$ is

Let $a$ and $b$ be unit vectors with $\theta$ as the acute angle between them. If $\frac{1}{2}|a-b|=\sin(\lambda \theta)$,then $4 \lambda^2=$

$A$ vector $\bar{a}$ has components $1$ and $2p$ with respect to a rectangular Cartesian system. This system is rotated through a certain angle about the origin in the counter-clockwise sense. If,with respect to the new system,$\bar{a}$ has components $1$ and $(p+1)$,then:

Let $u, v, w$ be vectors such that $u + v + w = 0$. If $|u| = 3, |v| = 4,$ and $|w| = 5,$ find the value of $u \cdot v + v \cdot w + w \cdot u$.