If $|\bar{a} \times \bar{b}|^2+(\bar{a} \cdot \bar{b})^2=144$ and $|\bar{a}|=4$,then $|\bar{b}|=$

If $\vec{a}$ and $\vec{b}$ are unit vectors,then the maximum value of $|\vec{a} + \vec{b}| + |\vec{a} - \vec{b}|$ is:

If $\triangle ABC$ is a right-angled triangle in which $BC$ is the hypotenuse, and the position vectors of $B$ and $C$ are $\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}$ and $\vec{c} = 5\hat{i} + \hat{j} - 3\hat{k}$ respectively, then the value of $\vec{AB} \cdot \vec{AC} + \vec{BA} \cdot \vec{BC} + \vec{CA} \cdot \vec{CB}$ is:

If $\vec{a}, \vec{b}, \vec{c}$ are three non-zero,non-coplanar vectors and $\vec{b_1} = \vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a}$,$\vec{b_2} = \vec{b} + \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a}$,and $\vec{c_1} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} + \frac{\vec{c} \cdot \vec{b}}{|\vec{b}|^2}\vec{b_1}$,$\vec{c_2} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c} \cdot \vec{b_1}}{|\vec{b_1}|^2}\vec{b_1}$,$\vec{c_3} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{c}|^2}\vec{a} + \frac{\vec{c} \cdot \vec{b_2}}{|\vec{c}|^2}\vec{b_1}$,$\vec{c_4} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{c}|^2}\vec{a} - \frac{\vec{b} \cdot \vec{c}}{|\vec{b}|^2}\vec{b_1}$. Then,which of the following is a set of mutually orthogonal vectors?

Let $\vec{a}=2 \hat{i}+5 \hat{j}-\hat{k}$,$\vec{b}=2 \hat{i}-2 \hat{j}+2 \hat{k}$ and $\vec{c}$ be three vectors such that $(\vec{c}+\hat{i}) \times (\vec{a}+\vec{b}+\hat{i}) = \vec{a} \times (\vec{c}+\hat{i})$ and $\vec{a} \cdot \vec{c} = -29$. Then $\vec{c} \cdot (-2 \hat{i}+\hat{j}+\hat{k})$ is equal to:

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