Find the magnitude of two vectors $\vec{a}$ and $\vec{b}$,having the same magnitude and such that the angle between them is $60^{\circ}$ and their scalar product is $\frac{1}{2}$.

In a triangle $ABC$ with usual notations,if $|\overline{BC}|=8, |\overline{CA}|=7, |\overline{AB}|=10$,then the projection of $\overline{AB}$ on $\overline{AC}$ is

$a, b$ and $c$ are three vectors with magnitudes $|a| = 4, |b| = 4, |c| = 2$ such that $a$ is perpendicular to $(b + c)$,$b$ is perpendicular to $(c + a)$,and $c$ is perpendicular to $(a + b)$. It follows that $|a + b + c|$ is equal to

If $a$ is a vector of magnitude $7$ and $b$ is a vector of magnitude $8$,then the maximum value of $|a \cdot b|$ is:

Let $(x, y) \in (R \times R)$ and $\vec{a} = x \hat{i} + 2 \hat{j} - \hat{k}$, $\vec{b} = 6 \hat{i} - y \hat{j} + 2 \hat{k}$ be two vectors. If $|\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = f(x) g(y)$, then $f(x) + g(y) - 46 = 0$ represents:

If $x$ and $y$ are two unit vectors and the angle between them is $\phi$,then $\frac{1}{2} |x - y| = $