JEE Main 2018 — Vectors & 3D Geometry

If the angle between the lines $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\frac{5-x}{-2}=\frac{7y-14}{P}=\frac{z-3}{4}$ is ${\mathrm{cos}}^{-1}(\frac{2}{3}),$ then $P$ is equal to

If the position vectors of the vertices $A, B$ and $C$ of a $\triangle \mathrm{ABC}$ are respectively $4 \hat{i}+7 \hat{j}+8 \hat{k}, 2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $2 \hat{i}+5 \hat{j}+7 \hat{k}$, then the position vector of the point, where the bisector of $\angle A$ meets $B C$ is

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k},\vec{c}=\hat{j}-\hat{k}$ and a vector $\vec{b}$ be such that $\vec{a}\times \vec{b}=\vec{c}$ and $\vec{a}\cdot \vec{b}=3$. Then $|\vec{b}|$ equals

Let $\vec{u}$ be a vector coplanar with the vectors $\vec{a}=2\hat{i}+3\hat{j}-\hat{k}$ and $\vec{b}= \hat{j}+\hat{k}$ . If $\vec{u}$ is perpendicular to $\vec{a}$ and $\vec{u}\cdot \vec{b}=24$, then ${|\vec{u}|}^{2}$ is equal to:

The sum of the intercepts on the coordinate axes of the plane passing through the point $(–2,–2,2)$ and containing the line joining the points $(1,–1,2)$ and $(1,1,1)$ is