If $\bar{a}, \bar{b}, \bar{c}, \bar{d}$ are unit vectors such that $\bar{a} \cdot \bar{b} = \frac{1}{2}$,$\bar{c} \cdot \bar{d} = \frac{1}{2}$ and the angle between $\bar{a} \times \bar{b}$ and $\bar{c} \times \bar{d}$ is $\frac{\pi}{6}$,then the value of $|[\bar{a} \bar{b} \bar{d}] \bar{c} - [\bar{a} \bar{b} \bar{c}] \bar{d}| = $

If $\vec{a}+\vec{b}+\vec{c}=0,$ show that $\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a} .$ Interpret the result geometrically.

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}+\hat{k}$ are two vectors,and $\vec{c}$ is a unit vector lying in the plane of $\vec{a}$ and $\vec{b}$ such that $\vec{c}$ is perpendicular to $\vec{b}$,then find the value of $\vec{c} \cdot (\hat{i}+\hat{j}+2\hat{k})$.

Let $\overrightarrow{a}=3 \hat{i}+2 \hat{j}+\hat{k}$,$\overrightarrow{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\overrightarrow{c}$ be a vector such that $(\vec{a}+\vec{b}) \times \vec{c}=2(\vec{a} \times \vec{b})+24 \hat{j}-6 \hat{k}$ and $(\overrightarrow{a}-\overrightarrow{b}+\hat{i}) \cdot \overrightarrow{c}=-3$. Then $|\overrightarrow{c}|^2$ is equal to . . . . . . .

Let $\vec{a}=2 \hat{i}+\alpha \hat{j}+\hat{k}$,$\vec{b}=-\hat{i}+\hat{k}$,and $\vec{c}=\beta \hat{j}-\hat{k}$,where $\alpha$ and $\beta$ are integers and $\alpha \beta=-6$. Let the values of the ordered pair $(\alpha, \beta)$ for which the area of the parallelogram with diagonals $\vec{a}+\vec{b}$ and $\vec{b}+\vec{c}$ is $\frac{\sqrt{21}}{2}$,be $(\alpha_1, \beta_1)$ and $(\alpha_2, \beta_2)$. Then $\alpha_1^2+\beta_1^2-\alpha_2 \beta_2$ is equal to

Let $\hat{a}$ be a unit vector perpendicular to the vectors $\overrightarrow{b} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\overrightarrow{c} = 2\hat{i} + 3\hat{j} - \hat{k}$,and makes an angle of $\cos^{-1}\left(-\frac{1}{3}\right)$ with the vector $\hat{i} + \hat{j} + \hat{k}$. If $\hat{a}$ makes an angle of $\frac{\pi}{3}$ with the vector $\hat{i} + \alpha\hat{j} + \hat{k}$,then the value of $\alpha$ is: