JEE Main 2015 — Vectors & 3D Geometry

If the shortest distance between the line $\frac{x-1}{\alpha }=\frac{y+1}{-1}=\frac{z}{1},(\alpha \neq -1)$, and $x+y+z+1=0=2x-y+z+3$ is $\frac{1}{\sqrt{3}}$,then value of $\alpha$ is :

In a parallelogram $ABCD, |\vec{AB}|=a, |\vec{AD}|=b& |\vec{AC}|=c$. $\vec{DB}\cdot \vec{AB}$ has the value:

Let $\vec{a}$and $\vec{b}$be two unit vectors such that $|\vec{a}+\vec{b}|=\sqrt{3}$. If $\vec{c}=\vec{a}+2\vec{b}+(\vec{a}\times \vec{b})$ , then $2|\vec{c}|$ is equal to:

The shortest distance between the $z$ - axis and the line $x+y+2z-3=0=2x+3y+4z-4,$ is

Let $\vec{a},\vec{b}\text{and}\vec{c}$ be three non - zero vectors such that no two of them are collinear and $(\vec{a}\times \vec{b})\times \vec{c}=\frac{1}{3}|\vec{b}||\vec{c}|\vec{a}$. If $\theta$ is the angle between vectors $\vec{b}\text{ and}\vec{c}$, then a value of $\mathrm{sin}\theta$ is