If $\bar{a}$ and $\bar{b}$ are unit vectors such that $|\bar{a}+\bar{b}|=\sqrt{3}$,then the angle between $\bar{a}$ and $\bar{b}$ is

Let $\vec{a}=2\hat{i}+\hat{j}-2\hat{k}$, $\vec{b}=\hat{i}+\hat{j}$ and $\vec{c}=\vec{a}\times\vec{b}$. Let $\vec{d}$ be a vector such that $|\vec{d}-\vec{a}|=\sqrt{11}$, $|\vec{c}\times\vec{d}|=3$ and the angle between $\vec{c}$ and $\vec{d}$ is $\frac{\pi}{4}$. Then $\vec{a}\cdot\vec{d}$ is equal to

If $\overrightarrow{a}=\hat{i}-\hat{j}-\hat{k}$ and $\overrightarrow{b}=\lambda \hat{i}-3 \hat{j}+\hat{k}$ and the orthogonal projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ is $\frac{4}{3}(\hat{i}-\hat{j}-\hat{k})$,then $\lambda$ is equal to

$A$ particle acted on by two forces $3i + 2j - 3k$ and $2i + 4j + 2k$ is displaced from the point $i + 2j + k$ to $5i + 4j + 2k.$ The total work done by the forces is equal to ............ $unit$.

If the vectors $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$,$\vec{b} = 2\hat{i} + 4\hat{j} + \hat{k}$,and $\vec{c} = \alpha\hat{i} + \hat{j} + \beta\hat{k}$ are mutually orthogonal,then $(\alpha, \beta) = $

The component of $\hat{i}$ in the direction of the vector $\hat{i}+\hat{j}+2 \hat{k}$ is