The work done by the force $F = 2i - 3j + 2k$ in displacing a particle from the point $(3, 4, 5)$ to the point $(1, 2, 3)$ is ............ $unit$.

If $a(\vec{\alpha} \times \vec{\beta}) + b(\vec{\beta} \times \vec{\gamma}) + c(\vec{\gamma} \times \vec{\alpha}) = \overrightarrow{0}$, where $a, b, c$ are non-zero scalars, then the vectors $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are

If $3\vec{a} - 5\vec{b}$ and $2\vec{a} + \vec{b}$ are perpendicular to each other,and $\vec{a} + 4\vec{b}$ and $-\vec{a} + \vec{b}$ are also perpendicular to each other,and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$,then find $\cos \theta$.

If $a \times r = b + \lambda a$ and $a \cdot r = 3,$ where $a = 2i + j - k$ and $b = -i - 2j + k,$ then $r$ and $\lambda$ are equal to

If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors and $\vec{a}+\vec{b}+\vec{c}=\vec{0}$,then the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ is . . . . . . . (in $/2$)

For non-zero vectors $a$ and $b$, if $|a+b| < |a-b|$, then $a$ and $b$ are