JEE Main 2006 — Vectors & 3D Geometry

A particle has two velocities of equal magnitude inclined to each other at an angle $\theta$. If one of them is halved, the angle between the other and the original resultant velocity is bisected by the new resultant. Then $\theta$ is

If $(\mathrm{a} \times \mathrm{b}) \times \overline{\mathrm{c}}=\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})$, where $\overline{\mathrm{a}}, \overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ are any three vectors such that $\overline{\mathrm{a}} \cdot \overline{\mathrm{b}} \neq 0$, $\overline{\mathrm{b}} \cdot \overline{\mathrm{c}} \neq 0$, then $\mathrm{a}$ and $\mathrm{c}$ are

$A B C$ is a triangle, right angled at $A$. The resultant of the forces acting along $\overrightarrow{A B}, \overrightarrow{A C}$ with magnitudes $\frac{1}{A B}$ and $\frac{1}{A C}$ respectively is the force along $\overrightarrow{A D}$, where $D$ is the foot of the perpendicular from $\mathrm{A}$ onto $\mathrm{BC}$. The magnitude of the resultant is

The values of $a$, for which the points $A, B, C$ with position vectors $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $a \hat{i}-3 \hat{j}+\hat{k}$ respectively are the vertices of a right-angled triangle with $C=\frac{\pi}{2}$ are