JEE Main 2013 — Vectors & 3D Geometry

A vector $\vec{n}$ is inclined to $x$-axis at $45^{\circ}$, to $y$-axis at $60^{\circ}$ and at an acute angle to $z$-axis. If $\vec{n}$ is a normal to a plane passing through the point $(\sqrt{2},-1,1)$ then the equation of the plane is :

If $\vec{a}$ and $\vec{b}$ are non-collinear vectors, then the value of $\alpha$ for which the vectors $\vec{u}=(\alpha-2) \vec{a}+\vec{b}$ and $\vec{v}=(2+3 \alpha) \vec{a}-3 \vec{b}$ are collinear is :

If $\hat{a}, \hat{b}$ and $\hat{c}$ are unit vectors satisfying $\hat{a}-\sqrt{3} \hat{b}+\hat{c}=\overrightarrow{0}$, then the angle between the vectors $\hat{a}$ and $\hat{c}$ is :

If the lines $\frac{x+1}{2}=\frac{y-1}{1}=\frac{z+1}{3}$ and $\frac{x+2}{2}=\frac{y-k}{3}=\frac{z}{4}$ are coplanar, then the value of $k$ is :

If the vectors $\vec{\mathrm{AB}}=3\hat{i}+4\hat{k}$ and $\vec{\mathrm{AC}}=5\hat{i}-2\hat{j}+4\hat{k}$ are the sides of a triangle $ABC,$ then the length of the median through $A$ is:

Let $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}-\hat{k}$ and $\vec{c}=\hat{i}+\hat{j}-2 \hat{k}$ be three vectors. A vector of the type $\vec{b}+\lambda \vec{c}$ for some scalar $\lambda$, whose projection on $\vec{a}$ is of magnitude $\sqrt{\frac{2}{3}}$ is :

Let $\mathrm{ABC}$ be a triangle with vertices at points $\mathrm{A}$ $(2,3,5)$, B $(-1,3,2)$ and $\mathrm{C}(\lambda, 5, \mu)$ in three dimensional space. If the median through $\mathrm{A}$ is equally inclined with the axes, then $(\lambda, \mu)$ is equal to:

Let $\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}, \vec{b}=\hat{i}+\hat{j}$. If $\vec{c}$ is a vector such that $\vec{a} \bullet \vec{c}=|\vec{c}|,|\vec{c}-\vec{a}|=2 \sqrt{2}$ and the angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is $30^{\circ}$, then $|(\vec{a} \times \vec{b}) \times \vec{c}|$ equals:

The vector $(\hat{i} \times \vec{a} \cdot \vec{b}) \hat{i}+(\hat{j} \times \vec{a} \vec{b}) \hat{j}+(\hat{k} \times \vec{a} \cdot \vec{b}) \hat{k}$ is equal to: