For non-zero vectors $a$ and $b$, if $|a+b| < |a-b|$, then $a$ and $b$ are

If $A, B, C$ and $D$ are four points in the plane such that $|AB|^2+|CD|^2=|BC|^2+|DA|^2=100$,then $AC \cdot BD=$

Let $\overline{u}, \overline{v}, \overline{w}$ be vectors such that $|\overline{u}|=1, |\overline{v}|=2, |\overline{w}|=3$. If the projection of $\overline{v}$ along $\overline{u}$ is equal to that of $\overline{w}$ along $\overline{u}$,and the vectors $\overline{v}$ and $\overline{w}$ are perpendicular to each other,then $|\overline{u}-\overline{v}+\overline{w}|$ equals:

The set of real values of $\lambda$ for which the vectors $\lambda \hat{i}-3 \hat{j}+5 \hat{k}$ and $2 \lambda \hat{i}-\lambda \hat{j}+\hat{k}$ are perpendicular to each other is

The angle between two consecutive sides $\vec{a}$ and $\vec{b}$ of a parallelogram is $\frac{\pi}{6}$. Given $\vec{a} = (2, -2, 1)$ and $\vec{b} = 2|\vec{a}|$,the area of the parallelogram is . . . . . . sq. units.

If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are three vectors such that $\overrightarrow{a}=\overrightarrow{b}+\overrightarrow{c}$ and the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is $\frac{\pi}{2}$, then: