Let $(x, y) \in (R \times R)$ and $\vec{a} = x \hat{i} + 2 \hat{j} - \hat{k}$, $\vec{b} = 6 \hat{i} - y \hat{j} + 2 \hat{k}$ be two vectors. If $|\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = f(x) g(y)$, then $f(x) + g(y) - 46 = 0$ represents:

Let $\vec{a}=6 \hat{i}+9 \hat{j}+12 \hat{k}$,$\vec{b}=\alpha \hat{i}+11 \hat{j}-2 \hat{k}$ and $\vec{c}$ be vectors such that $\vec{a} \times \vec{c}=\vec{a} \times \vec{b}$. If $\vec{a} \cdot \vec{c}=-12$ and $\vec{c} \cdot (\hat{i}-2 \hat{j}+\hat{k})=5$,then $\vec{c} \cdot (\hat{i}+\hat{j}+\hat{k})$ is equal to $.............$.

If $a, b$ and $c$ are unit vectors such that $a+b+c=0$,then $a \cdot b+b \cdot c+c \cdot a$ is equal to

Let $x \in R$ and $\log_2 x > 0$. Then,the vectors $A = (2, \log_2 x, s)$ and $B = (\log_2 x, s, \log_2 x)$ include an acute angle if

If $a = i + j + k$,$a \cdot b = 1$ and $a \times b = j - k$,then $b = $

In a parallelogram $OACB$,$\overrightarrow{OA} = \vec{a}$,$\overrightarrow{OB} = \vec{b}$,and the foot of the perpendicular drawn from point $B$ to $AC$ is $M$. If $\vec{a} \cdot \vec{b} = 1$ and $|\vec{a}| = |\vec{b}| = 2$,then $|\overrightarrow{BM}|$ is: