JEE Main 2016 — Vectors & 3D Geometry

In a triangle $ABC$, right angle at vertex $A$, if the position vectors of $A,B$ and $C$ are respectively $3\hat{i}+ \hat{j}- \hat{k}, -\hat{i}+3\hat{j}+p\hat{k}$ and $5\hat{i}+q\hat{j}-4\hat{k}$ , then the point $(p,q)$ lies on a line:

$ABC$ is a triangle in a plane with vertices $A(2, 3, 5), B(-1, 3, 2)$ and $C(\lambda , 5, \mu )$ . If the median through $A$ is equally inclined to the coordinate axes, then the value of $({\lambda }^{3}+{\mu }^{3}+5)$ is

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three unit vectors such that $\vec{a} \times (\vec{b} \times \vec{c})=\frac{\sqrt{3}}{2}(\vec{b} + \vec{c}).$ If $\vec{b}$ is not parallel to $\vec{c}$ , then the angle between $\vec{a}$ and $\vec{b}$ is

Let $ABC$ be a triangle whose circumcentre is at $P$. If the position vectors $A,B,C$ and $P$ are $\vec{a},\vec{b},\vec{c}$ and $\frac{\vec{a}+\vec{b}+\vec{c}}{4}$ respectively, then the position vector of the orthocentre of this triangle, is :

The number of distinct real values of $\lambda$, for which the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{{\lambda }^{2}}$ and $\frac{x-3}{1}=\frac{y-2}{{\lambda }^{2}}=\frac{z-1}{2}$ , are coplanar is

The shortest distance between the lines $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4},$ lies in the interval: