Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors such that $2 \vec{a}+3 \vec{b}+4 \vec{c}=\vec{0}$. Then $|\vec{b} \times \vec{c}|=$

Given $ABC$ is an equilateral triangle of side length $1$ unit and $P$ be any arbitrary point on the circumcircle of triangle $ABC,$ then $|\vec{PA}|^2+|\vec{PB}|^2+|\vec{PC}|^2$ is equal to

If $\vec{a} = -4 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and $\vec{b} = \sqrt{2} \hat{i} - \sqrt{2} \hat{j}$ are two vectors,then the angle between the vectors $2 \vec{a}$ and $\frac{\vec{b}}{2}$ is (in $^{\circ}$)

If $a, b, c$ and $r$ are vectors such that $a$ is not perpendicular to $b$,$r \times b = c \times b$ and $r \cdot a = 0$,then $r =$

Let $a, b, c \in \mathbb{R}$ be such that $a^{2} + b^{2} + c^{2} = 1$. If $a \cos \theta = b \cos \left(\theta + \frac{2\pi}{3}\right) = c \cos \left(\theta + \frac{4\pi}{3}\right)$ where $\theta = \frac{\pi}{9}$,then the angle between the vectors $\vec{p} = a \hat{i} + b \hat{j} + c \hat{k}$ and $\vec{q} = b \hat{i} + c \hat{j} + a \hat{k}$ is:

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