Q.If $\vec a=3\hat i-2\hat j+\hat k$ and $\vec b=2\hat i+\hat j-3\hat k$, find $\vec a+\vec b$ and $\vec a-\vec b$.

Place the regular hexagon's vertices as position vectors and compute each vector directly.

Let the regular hexagon $ABCDEF$ (side $s$) be centred at the origin with vertices placed symmetrically (each vertex at distance $s$ from the centre, $60^\circ$ apart):

$A=(-s,0),\ B=\left(-\tfrac{s}{2},-\tfrac{s\sqrt3}{2}\right),\ C=\left(\tfrac{s}{2},-\tfrac{s\sqrt3}{2}\right),\ D=(s,0),\ E=\left(\tfrac{s}{2},\tfrac{s\sqrt3}{2}\right),\ F=\left(-\tfrac{s}{2},\tfrac{s\sqrt3}{2}\right)$

Compute the required vectors:

$\vec{AB}=B-A=\left(\tfrac{s}{2},-\tfrac{s\sqrt3}{2}\right)$

$\vec{AD}=D-A=(2s,0)$

$\vec{EB}=B-E=(-s,-s\sqrt3)$

$\vec{FC}=C-F=(s,-s\sqrt3)$

Add them:

$\vec{AD}+\vec{EB}+\vec{FC} = (2s-s+s,\ 0-s\sqrt3-s\sqrt3) = (2s,-2s\sqrt3)$

Compare with $4\vec{AB} = 4\left(\tfrac{s}{2},-\tfrac{s\sqrt3}{2}\right) = (2s,-2s\sqrt3)$ — identical.

Hence $\vec{AD}+\vec{EB}+\vec{FC} = 4\vec{AB}$.

Find the unit vector along $\bar a$, then scale it to magnitude 7.

$\bar a = \bar i - 2\bar j$

$|\bar a| = \sqrt{1^2+(-2)^2} = \sqrt{5}$

Unit vector along $\bar a$: $\hat a = \dfrac{\bar a}{|\bar a|} = \dfrac{\bar i - 2\bar j}{\sqrt5}$

Required vector of magnitude 7 in the direction of $\bar a$:

$7\hat a = \dfrac{7}{\sqrt5}(\bar i - 2\bar j) = \dfrac{7}{\sqrt5}\bar i - \dfrac{14}{\sqrt5}\bar j = \dfrac{7\sqrt5}{5}\bar i - \dfrac{14\sqrt5}{5}\bar j$

Chain the given vectors head-to-tail: $\overrightarrow{OD} = \overrightarrow{OA}+\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}$.

Step 1 — Use the triangle/polygon law of vector addition.

Since $O\to A\to B\to C\to D$ is a chain of consecutive vectors,

$\overrightarrow{OD} = \overrightarrow{OA}+\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}$

Step 2 — Add component-wise.

$\bar i$: $1+3+1+2 = 7$

$\bar j$: $1-2+2+1 = 2$

$\bar k$: $1+1-2+3 = 3$

Step 3 — Write the resultant.

$\overrightarrow{OD} = 7\bar i + 2\bar j + 3\bar k$

Add the three vectors, find the magnitude of the sum, then negate the unit vector to get the opposite direction.

Given $\bar{a} = 2\bar{i}+4\bar{j}-5\bar{k}$, $\bar{b} = \bar{i}+\bar{j}+\bar{k}$, $\bar{c} = \bar{j}+2\bar{k}$.

$\bar{a}+\bar{b}+\bar{c} = (2+1+0)\bar{i} + (4+1+1)\bar{j} + (-5+1+2)\bar{k} = 3\bar{i}+6\bar{j}-2\bar{k}$

Magnitude: $|\bar{a}+\bar{b}+\bar{c}| = \sqrt{3^2+6^2+(-2)^2} = \sqrt{9+36+4} = \sqrt{49} = 7$

Unit vector in the same direction: $\dfrac{3\bar{i}+6\bar{j}-2\bar{k}}{7}$

Unit vector in the opposite direction: negate it.