If the $p^{th}$,$q^{th}$,and $r^{th}$ terms of a geometric progression are $a$,$b$,and $c$ respectively,then find the angle between the vectors $\vec{u} = (\log a)\hat{i} + (\log b)\hat{j} + (\log c)\hat{k}$ and $\vec{v} = (q - r)\hat{i} + (r - p)\hat{j} + (p - q)\hat{k}$.

If the constant forces $2 \hat{i}-5 \hat{j}+6 \hat{k}$ and $-\hat{i}+2 \hat{j}-\hat{k}$ act on a particle due to which it is displaced from a point $A(4,-3,-2)$ to a point $B(6,1,-3)$, then the work done by the forces is (in $\text{ unit}$)

Let $\vec{a}=\alpha \hat{i}+\hat{j}+\beta \hat{k}$ and $\vec{b}=3 \hat{i}-5 \hat{j}+4 \hat{k}$ be two vectors,such that $\vec{a} \times \vec{b}=-\hat{i}+9 \hat{j}+12 \hat{k}$. Then the projection of $\vec{b}-2 \vec{a}$ on $\vec{b}+\vec{a}$ is equal to.

Let $\lambda \in \mathbb{Z}$,$\vec{a} = \lambda \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 3 \hat{i} - \hat{j} + 2 \hat{k}$. Let $\vec{c}$ be a vector such that $(\vec{a} + \vec{b} + \vec{c}) \times \vec{c} = \vec{0}$,$\vec{a} \cdot \vec{c} = -17$ and $\vec{b} \cdot \vec{c} = -20$. Then $|\vec{c} \times (\lambda \hat{i} + \hat{j} + \hat{k})|^2$ is equal to

The vectors $2\,i + 3\,j - 4\,k$ and $a\,i + b\,j + c\,k$ are perpendicular,when

If $\vec{a}, \vec{b}, \vec{c}$ are vectors such that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ and $|\vec{a}|=7, |\vec{b}|=5, |\vec{c}|=3$,then the angle between vector $\vec{b}$ and $\vec{c}$ is: (in $^{\circ}$)