Let $u, v$ and $w$ be three vectors such that $u+v+w=0$,$|u|=3$,$|v|=5$ and $|w|=7$. Then the angle between $u$ and $v$ is (in $^{\circ}$)

Let $\vec{a}=2 \hat{i}+3 \hat{j}+\hat{k}$,$\vec{b}=4 \hat{i}+\hat{j}$,$\vec{c}=\hat{i}-3 \hat{j}-7 \hat{k}$. If $\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$,$\vec{r} \cdot \vec{a}=9$,$\vec{r} \cdot \vec{b}=7$,$\vec{r} \cdot \vec{c}=6$,then $(x, y, z) = $

If $\bar{a} = \hat{i} + \hat{j}$ and $\bar{b} = 2\hat{i} - \hat{k}$,then the point of intersection of the lines $\bar{r} \times \bar{a} = \bar{b} \times \bar{a}$ and $\bar{r} \times \bar{b} = \bar{a} \times \bar{b}$ is

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 $\text{ unit}$)

The projection of $\overrightarrow{a} = 3 \hat{i} - \hat{j} + 5 \hat{k}$ on $\overrightarrow{b} = 2 \hat{i} + 3 \hat{j} + \hat{k}$ is

Let $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$. If $\vec{b}$ is a vector such that $\vec{a} \cdot \vec{b} = |\vec{b}|^2$ and $|\vec{a} - \vec{b}| = \sqrt{7}$,then find $|\vec{b}|$.