If points $D, E, F$ divide the sides $BC, CA, AB$ of $\triangle ABC$ in the ratios $1:4, 3:2, 3:7$ respectively,and point $K$ divides $AB$ in some ratio,then $(\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF}) : \overrightarrow{CK} = ......$

If $\bar{a}, \bar{b}, \bar{c}$ are three vectors such that $|\bar{a}|=\sqrt{31}, 4|\bar{b}|=|\bar{c}|=2$ and $2(\bar{a} \times \bar{b})=3(\bar{c} \times \bar{a})$ and if the angle between $\bar{b}$ and $\bar{c}$ is $\frac{2\pi}{3}$,then $\left|\frac{\bar{a} \times \bar{c}}{\bar{a} \cdot \bar{b}}\right|^2=$

Let $\vec{a}_n = (\tan \theta_n)\hat{i} + \hat{j}$ and $\vec{b}_n = \hat{i} - (\cot \theta_n)\hat{j}$, where $\theta_n = \frac{2^{n-1}\pi}{2^n+1}$, for some $n \in N, n > 5$. Then the value of $\frac{\sum_{k=1}^n |\vec{a}_k|^2}{\sum_{k=1}^n |\vec{b}_k|^2}$ is . . . . . . .

Given that $a$ and $b$ are two unit non-collinear vectors,if $u = a - (a \cdot b)b$ and $v = a \times b$,then find $|v| =$.

Let $\hat{i}, \hat{j}$ and $\hat{k}$ be the unit vectors along the three positive coordinate axes. Let $\vec{a}=3\hat{i}+\hat{j}-\hat{k}$,$\vec{b}=\hat{i}+b_2\hat{j}+b_3\hat{k}$ $(b_2, b_3 \in \mathbb{R})$,and $\vec{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$ $(c_1, c_2, c_3 \in \mathbb{R})$ be three vectors such that $b_2b_3 > 0$,$\vec{a} \cdot \vec{b} = 0$ and $\begin{bmatrix} 0 & -c_3 & c_2 \\ c_3 & 0 & -c_1 \\ -c_2 & c_1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ b_2 \\ b_3 \end{bmatrix} = \begin{bmatrix} 3-c_1 \\ 1-c_2 \\ -1-c_3 \end{bmatrix}$. Then,which of the following is/are $TRUE$?

In the given figure,if a vector $x$ satisfies the equation $x - w = v$,then $x = ?$