If $a, b, c$ are unit vectors satisfying the relation $a+b+\sqrt{3} c=0$, then the angle between $a$ and $b$ is

If $(a \times b)^{2} + (a \cdot b)^{2} = 144$ and $|a| = 4$,then $|b|$ is equal to

Let $a = i + 2j + k$, $b = i - j + k$, and $c = i + j - k$. $A$ vector in the plane of $a$ and $b$ has a projection of $\frac{1}{\sqrt{3}}$ on $c$. Then, one such vector is:

$A$ vector $\overrightarrow{a} = \alpha \hat{i} + 2 \hat{j} + \beta \hat{k}$ (where $\alpha, \beta \in R$) lies in the plane of the vectors $\overrightarrow{b} = \hat{i} + \hat{j}$ and $\overrightarrow{c} = \hat{i} - \hat{j} + 4 \hat{k}$. If $\overrightarrow{a}$ bisects the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$,then:

If $\vec{a} = 4 \hat{i} + 5 \hat{j} - 3 \hat{k}$ and $\vec{b} = 6 \hat{i} - 2 \hat{j} - 2 \hat{k}$ are two vectors, then the magnitude of the component of $\vec{b}$ parallel to $\vec{a}$ is: (in $\sqrt{2}$)

The angle between the vectors $\vec{a} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$ is