If $a \cdot i = 4$,then $(a \times j) \cdot (2j - 3k) = $

The angle between two diagonals of a cube is:

The position vectors of the vertices of a triangle $ABC$ are $4i - 2j$,$i + 4j - 3k$,and $-i + 5j + k$ respectively. Then $\angle ABC = $

If $\vec{a} = 4\hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{c} = 3\hat{i} + 4\hat{j} + 5\hat{k}$, then $(\vec{a} + \vec{b}) \cdot (\vec{b} + \vec{c}) =$

If vectors $\vec{a}$ and $\vec{b}$ are not perpendicular,and $\vec{c}$ and $\vec{d}$ are two vectors satisfying $\vec{b} \times \vec{c} = \vec{b} \times \vec{d}$ and $\vec{a} \cdot \vec{d} = 0$,then the vector $\vec{d}$ is equal to:

Let three vectors $\overrightarrow{a}=\alpha \hat{i}+4 \hat{j}+2 \hat{k}$,$\vec{b}=5 \hat{i}+3 \hat{j}+4 \hat{k}$,and $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$ form a triangle such that $\overrightarrow{c}=\overrightarrow{a}-\overrightarrow{b}$ and the area of the triangle is $5 \sqrt{6}$. If $\alpha$ is a positive real number,then $|\overrightarrow{c}|^2$ is: