Establish the following vector inequalities geometrically or otherwise:

$(a)$ $\quad| a + b | \leq| a |+| b |$

$(b)$ $\quad| a + b | \geq| a |-| b |$

$(c)$ $\quad| a - b | \leq| a |+| b |$

$(d)$ $\quad| a - b | \geq| a |-| b |$

When does the equality sign above apply?

$(a)$ Let two vectors $\vec{a}$ and $\vec{b}$ be represented by the adjacent sides of a parallelogram $OMNP$, as shown in the given figure.

Here, we can write:

$|\overrightarrow{ OM }|=|\vec{a}|$

$|\overrightarrow{ MN }|=|\overrightarrow{ OP }|=|\vec{b}|$

$|\overrightarrow{ ON }|=|\vec{a}+\vec{b}|$

In a triangle, each side is smaller than the sum of the other two sides. Therefore, in $\Delta$ $OMN$, we have:

$ON \,<\, ( OM + MN )$

$|\vec{a}+\vec{b}| \,<\, |\vec{a}|+|\vec{b}|$

If the two vectors $\vec{a}$ and $\vec{b}$ act along a straight line in the same direction, then we can write:

$|\vec{a}+\vec{b}|=|\vec{a}|+|\vec{b}|$

Combining above equations we get:

$|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|$

$(b)$ Let two vectors $\vec{a}$ and $\vec{b}$ be represented by the adjacent sides of a parallelogram $OMNP$, as shown in the given figure.

In a triangle, each side is smaller than the sum of the other two sides. Therefore, in $\Delta$ $OMN$, we have

$ON + MN \,>\, OM$

$ON + OM \,>\, MN$

$|\overrightarrow{ ON }|\,>\,|\overrightarrow{ OM }-\overrightarrow{ OP }|$

$(\because OP = MN )$

$|\vec{a}+\vec{b}|\,>\,|| \vec{a}|-| \vec{b}||$

If the two vectors $\vec{a}$ and $\vec{b}$ act along a straight line in the same direction, then we can write:

$|\vec{a}+\vec{b}|=|| \vec{a}|-| \vec{b}||$

Combining above equations, we get:

$|\vec{a}+\vec{b}| \geq|| \vec{a}|-| \vec{b}||$

$(c)$ Let two vectors $\vec{a}$ and $\vec{b}$ be represented by the adjacent sides of a parallelogram $PORS$, as shown in the given figure.

Here we have:

$|\overrightarrow{ OR }|=|\overrightarrow{ PS }|=|\vec{b}|$

$|\overrightarrow{ OP }|=|\vec{a}|$

In a triangle, each side is smaller than the sum of the other two sides. Therefore, in $\Delta$ $OPS$, we have:

$OS \,<\, OP + PS$

$|\vec{a}-\vec{b}|  \,<\,  |\vec{a}|+|-\vec{b}|$

$|\vec{a}-\vec{b}|  \,<\,  |\vec{a}|+|\vec{b}|$

If the two vectors act in a straight line but in opposite directions, then we can write:

$|\vec{a}-\vec{b}|=|\vec{a}|+|\vec{b}|$

Combining above equations, we get:

$|\vec{a}-\vec{b}| \leq|\vec{a}|+|\vec{b}|$

$(d)$ Let two vectors $\vec{a}$ and $\vec{b}$ be represented by the adjacent sides of a parallelogram $PORS$, as shown in the given figure.

The following relations can be written for the given parallelogram. $OS + PS > OP$

$OS \,>\, OP-PS $

$|\vec{a}-\vec{b}|\,>\,|\vec{a}|-|\vec{b}|$

The quantity on the $LHS$ is always positive and that on the $RHS$ can be positive or negative. To make both quantities positive, we take modulus on both sides as:

||$\vec{a}-\vec{b}||\,>\,|| \vec{a}|-| \vec{b}||$

$|\vec{a}-\vec{b}|\,>\,|| \vec{a}|-| \vec{b}||$

If the two vectors act in a straight line but in the opposite directions, then we can write:

$|\vec{a}-\vec{b}|=|| \vec{a}|-| \vec{b}||$

Combining equations , we get:

$|\vec{a}-\vec{b}| \geq|| \vec{a}|-| \vec{b} |$