Q.Let $A,B,C$ be the vertices of a triangle. Let $D,E,F$ be the midpoints of the sides $BC,CA,AB$ respectively. Show that $\vec{AD}+\vec{BE}+\vec{CF}=\vec 0$.

Using the triangle law repeatedly, $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ and $\overrightarrow{CD}+\overrightarrow{DA}=\overrightarrow{CA}$, and these two cancel to give $\vec{0}$.

$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{DA}+\overrightarrow{CD}$

Group the first two: $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ (triangle law: consecutive vectors add head-to-tail).

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Two vectors expressed in the same basis $\vec a,\vec b$ are parallel exactly when their coefficients are proportional.

If $\vec a+2\vec b$ is parallel to $3\vec a+m\vec b$, then $3\vec a+m\vec b=k(\vec a+2\vec b)$ for some scalar $k$. …

Rearranging the four vectors as $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}$ traces a closed path $A\to B\to C\to D\to A$, which always sums to zero.

Vector addition does not depend on the order in which the vectors are added:

$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{DA}+\overrightarrow{CD} = \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}$

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With $\overrightarrow{AB}=\vec a$ and $\overrightarrow{AD}=\vec b$ as adjacent sides, the other diagonal is $\overrightarrow{BD}=\vec b-\vec a$.

In parallelogram $ABCD$, take $A$ as the origin. Then $\overrightarrow{AB}=\vec a$ and $\overrightarrow{AD}=\vec b$ are the adjacent sides, and the diagonal $\overrightarrow{AC}=\vec a+\vec b$ is given, consistent with $C = A+\vec a+\vec b$ (opposite vertex).

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The resultant vector is $2\hat i-\hat j$; dividing by its magnitude $\sqrt5$ gives the unit vector.

The resultant (sum) of the two vectors is found by adding corresponding components:

$(\hat i+\hat j-\hat k)+(\hat i-2\hat j+\hat k)=(1+1)\hat i+(1-2)\hat j+(-1+1)\hat k=2\hat i-\hat j+0\hat k=2\hat i-\hat j.$

Its magnitude is $\sqrt{2^2+(-1)^2}=\sqrt{4+1}=\sqrt5$.

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The resultant of $\hat{i}+\hat{j}-\hat{k}$ and $\hat{i}-2\hat{j}+\hat{k}$ is $2\hat{i}-\hat{j}$ (the $\hat{k}$ components cancel); dividing by its magnitude $\sqrt{5}$ gives the unit vector.

Resultant $= (\hat{i}+\hat{j}-\hat{k}) + (\hat{i}-2\hat{j}+\hat{k})$.

Add componentwise: $\hat{i}$-component: $1+1=2$; $\hat{j}$-component: $1-2=-1$; $\hat{k}$-component: $-1+1=0$.

Resultant $= 2\hat{i}-\hat{j}$.

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$|k\vec a|=|k||\vec a|=1$ gives $|k|=1/|\vec a|$, so $k$ can be either $+1/|\vec a|$ or $-1/|\vec a|$.

For a scalar $k$ and vector $\vec a$, $|k\vec a| = |k|\cdot|\vec a|$ (magnitude of a scaled vector is the absolute value of the scalar times the vector's magnitude).

Given $|k\vec a|=1$: $|k|\cdot|\vec a| = 1 \Rightarrow |k| = \dfrac{1}{|\vec a|} \Rightarrow k = \pm\dfrac{1}{|\vec a|}$

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