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

Let $\vec{a}$ and $\vec{b}$ be two unit vectors. If the vectors $\vec{c} = \vec{a} + 2\vec{b}$ and $\vec{d} = 5\vec{a} - 4\vec{b}$ are perpendicular to each other,then the angle between $\vec{a}$ and $\vec{b}$ is:

If the components of $\vec{a}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ along and perpendicular to $\vec{b}=3 \hat{i}+\hat{j}-\hat{k}$ respectively,are $\frac{16}{11}(3 \hat{i}+\hat{j}-\hat{k})$ and $\frac{1}{11}(-4 \hat{i}-5 \hat{j}-17 \hat{k})$,then $\alpha^2+\beta^2+\gamma^2$ is equal to :

If $\theta$ is the 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

Let three vectors $\overrightarrow{a}, \overrightarrow{b}$ and $\overrightarrow{c}$ be such that $\overrightarrow{c}$ is coplanar with $\overrightarrow{a}$ and $\overrightarrow{b}$,$\overrightarrow{a} \cdot \overrightarrow{c} = 7$ and $\overrightarrow{b}$ is perpendicular to $\overrightarrow{c}$,where $\overrightarrow{a} = -\hat{i} + \hat{j} + \hat{k}$ and $\overrightarrow{b} = 2\hat{i} + \hat{k}$. Then the value of $2|\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}|^{2}$ is .........

If three consecutive vertices of a parallelogram are $A(4,3,5)$,$B(0,6,0)$,$C(-8,1,4)$ and $D$ is the fourth vertex,then the angle between $AC$ and $BD$ is