JEE Main 2017 — Vectors & 3D Geometry

The area (in sq. units) of the parallelogram whose diagonals are along the vectors $8\hat{i}-6\hat{j}$ and $3\hat{i}+4\hat{j}-12\hat{k},$ is:

The distance of the point $(1, 3,-7)$ from the plane passing through the point $(1,-1,-1)$ , having normal perpendicular to both the lines $\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}$ and $\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}$ , is:

Given, $\vec{a}=2\hat{i}+\hat{j}-2\hat{k}$ and $\vec{b}= \hat{i}+\hat{j}.$ Let $\vec{c}$ be a vector such that $|\vec{c}- \vec{a}|=3, |(\vec{a}\times \vec{b})\times \vec{c}|=3$ and the angle between $\vec{c}$ and $\vec{a}\times \vec{b}$ be $30^{\circ}$ . Then $\vec{a}\cdot \vec{c}$ is equal to:

If the vector $\vec{b}=3\hat{j}+4\hat{k}$ is written as the sum of a vector $\vec{{b}_{1}}$, parallel to $\vec{a}= \hat{i}+ \hat{j}$ and a vector ${\vec{b}}_{2},$ perpendicular to $\vec{a},$ then ${\vec{b}}_{1}\times {\vec{b}}_{2}$ is equal to :