Establish the following vector inequalities geometrically or otherwise:
Establish the following vector inequalities geometrically or otherwise:

    \[\left| \vec{a}-\vec{b} \right|\ge \left| \left| {\vec{a}} \right|-\left| {\vec{b}} \right| \right|\]

Answer –

Let two vectors \vec{a}and \vec{b}represent two adjacent sides of a parallelogram PQRS, as shown in the diagram below :

Here,

\left| \overrightarrow{SR} \right|=|\vec{a}| and \left| \overrightarrow{PS} \right|=\left| -\vec{b} \right|=\left| \overrightarrow{b} \right|

Also, \left| \overrightarrow{QS} \right|=\left| \vec{a}-\vec{b} \right|

We know that for a triangle, each side is always smaller than the sum of other two sides. So, we can write

QS < SR + PS

    \[\left| \vec{a}-\vec{b} \right|>\left| {\vec{a}} \right|-\left| {\vec{b}} \right|\to (1)\]

The quantity on the left is always positive, but the quantity on the right might be either positive or negative. To make both quantities positive, we use modulus on both sides:

\left| \left| \overrightarrow{a}-\overrightarrow{b} \right| \right|>\left| \left| \overrightarrow{a} \right|-\left| \overrightarrow{b} \right| \right|

\left| \overrightarrow{a}-\overrightarrow{b} \right|>\left| \left| \overrightarrow{a} \right|-\left| \overrightarrow{b} \right| \right|\to (2)

 

Now, if both the said vectors act in a straight line in opposite direction, then we can write,

 

    \[\left| \vec{a}-\vec{b} \right|=\left| \left| {\vec{a}} \right|-\left| {\vec{b}} \right| \right|\to (3)\]

Upon combining (2) and (3), we get the required result as,

    \[\left| \vec{a}-\vec{b} \right|\ge \left| \left| {\vec{a}} \right|-\left| {\vec{b}} \right| \right|\]