Is it possible that $\vec{A} + \vec{B} = \vec{A} - \vec{B}$?

The direction cosines of vector $(A - B)$,if $A = 2\hat{i} + 3\hat{j} + \hat{k}$ and $B = 2\hat{i} + 2\hat{j} + 3\hat{k}$,are:

What is the minimum number of coplanar vectors having different magnitudes that can be added to give a zero resultant?

Let $\vec{P} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{Q} = -(\hat{i} + \hat{j} + \hat{k})$. The angle between $(\vec{P} - \vec{Q})$ and $\vec{P}$ is (in $^\circ$)

The resultant force of $5 \, N$ and $10 \, N$ cannot be ........ $N$.

Vectors $\vec{A}$ and $\vec{B}$ make angles of $20^\circ$ and $110^\circ$ with the $X$-axis,respectively. The magnitudes of these vectors are $5 \, m$ and $12 \, m$,respectively. The magnitude of the resultant vector of these two vectors is ....... $m$.