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 $\vec{a}$ and $\vec{b}$ are unit vectors,then what is the angle between $\vec{a}$ and $\vec{b}$ for $\sqrt{3}\vec{a} - \vec{b}$ to be a unit vector (in $^{\circ}$)?

If $12 \hat{i}-12 \hat{j}-18 \hat{k}$,$-3 \hat{i}-6 \hat{j}-9 \hat{k}$ and $3 \hat{i}+3 \hat{j}-24 \hat{k}$ are the position vectors of the vertices $A, B$ and $C$ respectively of $\triangle ABC$,then the position vector of the incentre of $\triangle ABC$ is

If the components of $\vec{a}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ along and perpendicular to $\vec{b}=3 \hat{i}+\hat{j}-\hat{k}$ respectively,are $\frac{16}{11}(3 \hat{i}+\hat{j}-\hat{k})$ and $\frac{1}{11}(-4 \hat{i}-5 \hat{j}-17 \hat{k})$,then $\alpha^2+\beta^2+\gamma^2$ is equal to :

Let $\vec{AB} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $\vec{AD} = \hat{i} + 2 \hat{j} + \lambda \hat{k}$, $\lambda \in R$. Let the projection of the vector $\vec{v} = \hat{i} + \hat{j} + \hat{k}$ on the diagonal $\vec{AC}$ of the parallelogram $ABCD$ be of length $1$ unit. If $\alpha, \beta$, where $\alpha > \beta$, are the roots of the equation $\lambda^2 x^2 - 6 \lambda x + 5 = 0$, then $2 \alpha - \beta$ is equal to

The angle between two diagonals of a cube is: