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

If $$\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$$ and $$\mathbf{c}=x \hat{\mathbf{i}}+(x-2) \hat{\mathbf{j}}-\hat{\mathbf{k}}$$ and if the vector $$\mathbf{c}$$ lies in the plane of vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ and then $$x$$ equals