Mathematics Vectors & 3D Geometry questions from JEE Main 2015.
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: