A vector $$\bar{a}$$ has components 1 and $$2 p$$ with respect to a rectangular Cartesian system. This system is rotated through a certain angle about origin in the counter clock wise sense. If, with respect to the new system, $$\bar{a}$$ has components 1 and $$(p+1)$$, then

If $$\theta$$ is angle between the vectors $$\bar{a}$$ and $$\bar{b}$$ where $$|\bar{a}|=4,|\bar{b}|=3$$ and $$\theta \in\left(\frac{\pi}{4}, \frac{\pi}{3}\right)$$, then $$|(\bar{a}-\bar{b}) \times(\bar{a}+\bar{b})|^2+4(\bar{a} \cdot \bar{b})^2$$ has the value

$$A, B, C, D$$ are four points in a plane with position vectors $$\bar{a}, \bar{b}, \bar{c}, \bar{d}$$ respectively such that $$(\bar{a}-\bar{d}) \cdot(\bar{b}-\bar{c})=(\bar{b}-\bar{d}) \cdot(\bar{c}-\bar{a})=0$$. The point $$D$$, then is the ___________ of $$\triangle \mathrm{ABC}$$

Two adjacent of sides parallelogram $$\mathrm{ABCD}$$ are given by $$\overline{\mathrm{AB}}=2 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}$$ and $$\overline{A D}=-\hat{i}+2 \hat{j}+2 \hat{k}$$. The side $$A D$$ is rotated by angle $$\alpha$$ in plane of parallelogram so that $$\mathrm{AD}$$ becomes $$\mathrm{AD}^{\prime}$$. If $$\mathrm{AD}^{\prime}$$ makes a right angle with the side $$A B$$, then the cosine of the angle $$\alpha$$ is given by