The area of the parallelogram determined by vectors $A = 2\hat{i} + \hat{j} - 3\hat{k}$ and $B = 12\hat{j} - 2\hat{k}$ is approximately:

Consider three vectors $A = \hat{i} + \hat{j} - 2\hat{k}$,$B = \hat{i} - \hat{j} + \hat{k}$,and $C = 2\hat{i} - 3\hat{j} + 4\hat{k}$. $A$ vector $X$ of the form $X = \alpha A + \beta B$ (where $\alpha$ and $\beta$ are scalars) is perpendicular to $C$. The ratio of $\alpha$ to $\beta$ is: (in $: 1$)

Find the unit vector perpendicular to the two vectors $\vec{A} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{B} = \hat{i} + \hat{j} + \hat{k}$.

Two vectors $a \hat{i} + b \hat{j} + \hat{k}$ and $2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ are perpendicular to each other. Given $3a + 2b = 7$,the ratio of $a$ to $b$ is $\frac{x}{2}$. The value of $x$ is:

The angle between the vectors $(\hat{i} + \hat{j})$ and $(\hat{i} - \hat{k})$ is ........ $^\circ$.

The angle between the two vectors $\vec A = 3\hat i + 4\hat j + 5\hat k$ and $\vec B = 3\hat i + 4\hat j - 5\hat k$ will be....... $^o$