In quadrilateral $$A B C D, \mathbf{A B}=\mathbf{a}, \mathbf{B C}=\mathbf{b}$$. $$\mathbf{D A}=\mathbf{a}-\mathbf{b}, M$$ is the mid-point of $$B C$$ and $$X$$ is a point on DM such that, $$\mathbf{D X}=\frac{4}{5}$$ DM. Then, the points $$A, X$$ and $$C$$.

The vectors $$3 \mathbf{a}-5 \mathbf{b}$$ and $$2 \mathbf{a}+\mathbf{b}$$ are mutually perpendicular and the vectors $$a+4 b$$ and $$-\mathbf{a}+\mathbf{b}$$ are also mutually perpendicular, then the acute angle between $$\mathbf{a}$$ and $$\mathbf{b}$$ is

Let $$\mathbf{a}=x \hat{i}+y \hat{j}+z \hat{k}$$ and $$x=2 y$$. If $$|\mathbf{a}|=5 \sqrt{2}$$ and a makes an angle of $$135^{\circ}$$ with the Z-axis, then $$\mathbf{a}=$$

Let $$\mathbf{a}, \mathbf{b}, \mathbf{c}$$ be the position vectors of the vertices of a $$\triangle A B C$$. Through the vertices, lines are drawn parallel to the sides to form the $$\Delta A^{\prime} B^{\prime} C^{\prime}$$. Then, the centroid of $$\Delta A^{\prime} B^{\prime} C^{\prime}$$ is