For any two vectors $\vec{a}$ and $\vec{b}$,we always have $|\vec{a} \cdot \vec{b}| \leq |\vec{a}| |\vec{b}|$ (Cauchy-Schwarz inequality). Is this statement true or false?

Let the unit vectors $a$ and $b$ be perpendicular and the unit vector $c$ be inclined at an angle $\theta$ to both $a$ and $b$. If $c = \alpha a + \beta b + \gamma (a \times b)$,then

The projection of $\bar{a} = \hat{i} - 2\hat{j} + \hat{k}$ on $\bar{b} = 2\hat{i} - \hat{j} + \hat{k}$ is

In a parallelogram $ABCD$,the position vectors of vertices $A$ and $C$ are $3\hat{i} + 3\hat{j} + 5\hat{k}$ and $\hat{i} - 5\hat{j} - 5\hat{k}$ respectively. If $M$ is the midpoint of the diagonal $DB$,find the projection of $\overline{OM}$ on $\overline{OC}$,where $O$ is the origin.

If $\vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$ are orthogonal,then the value of $\lambda$ is:

If $\overrightarrow{a}=\hat{i}-\hat{j}-\hat{k}$ and $\overrightarrow{b}=\lambda \hat{i}-3 \hat{j}+\hat{k}$ and the orthogonal projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ is $\frac{4}{3}(\hat{i}-\hat{j}-\hat{k})$,then $\lambda$ is equal to