JEE Main 2014 — Vectors & 3D Geometry

Equation of the line of the shortest distance between the lines $\frac{ x⁡ }{ 1 } = \frac{ y⁡ }{ - 1 } = \frac{ z⁡ }{ 1 }$ and $\frac{ x⁡ - 1 }{ 0 } = \frac{ y⁡ + 1 }{ - 2 } = \frac{ z⁡ }{ 1 }$ is

If $\hat{x}, \hat{y}$ and $\hat{z}$ are three unit vectors in threedimensional space, then the minimum value of $|\hat{x}+\hat{y}|^2+|\hat{y}+\hat{z}|^2+|\hat{z}+\hat{x}|^2$

If $|\vec{c}|^2=60$ and $\vec{c} \times(\hat{i}+2 \hat{j}+5 \hat{k})=\overrightarrow{0}$, then a value of $\vec{c} \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$ is:

If $|\vec{a}|=2,|\vec{b}|=3$ and $|2\vec{a}-\vec{b}|=5,$ then $|2\vec{a}+\vec{b}|$ equals :

If $\vec{x}=3\hat{i}-6\hat{j}-\hat{k},\vec{y}=\hat{i}+4\hat{j}-3\hat{k}$ and $\vec{z}=3\hat{i}-4\hat{j}-12\hat{k}$, then the magnitude of the projection of $\vec{x}\times \vec{y}$ on $\vec{z}$ is

The angle between the lines whose direction cosines satisfy the equations $l⁡+m+n=0$ and $l{⁡}^{2}={m}^{2}+{n}^{2}$ is