Q.If ABCDEF is a regular hexagon, then prove that AD⃗ + EB⃗ + FC⃗ = 4AB⃗.

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:

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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 $\bar a$:

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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.

$\bar i$: $1+3+1+2 = 7$ …

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

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

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

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