Consider that the two vectors $\vec{a}$ and $\vec{b}$ are represented by $\overline{OP}$ and $\overline{OQ}$ . The addition of the two vectors i.e. $\vec{a}+\vec{b}$ is given by $\overline{OR}$ asshown in figure (i) (a) To prove $\left|\vec{a}+\vec{b}\right|\le\left|\vec{a}\right|+\left|\vec{b}\right|$ In Figure (i) consider the ΔOQR. Sinceone side of a triangle is always smallerthan the sum of the other two sides, itfollows that OR< QR + OQ or OR < OP + OQ Now, $OR=\left|\vec{a}+\vec{b}\right|, OP=\left|\vec{a}\right|$ and $OQ=\left|\vec{b}\right|$ $\therefore \left|\vec{a}+\vec{b}\right| < \left|\vec{a}\right|+\left|\vec{b}\right|\dots$ (i) In case, the vectors $\vec{a}$ and $\vec{b}$ are along the same straight line and point inthe same direction, then $\left|\vec{a}+\vec{b}\right|=\left|\vec{a}\right|+\left|\vec{b}\right|\dots$ (ii) Combining the conditions stated in the equations (i) and (ii), we have $\left|\vec{a}+\vec{b}\right|\le\left|\vec{a}\right|+\left|\vec{b}\right|$ (b) To prove $\left|\vec{a}+\vec{b}\right|\ge\left|\left|\vec{a}\right|-\left|\vec{b}\right|\right|$ In Figure (i) again consider the ΔOQR. It follows that OR + OQ > OR or OR > |QR – OQ| The modulus of QR – OR has been taken for the reason that whereas theL.H.S. is always positive, the R.H.S. may be negative in case QR is smallerthan OQ. Since QR = OP, OR > |OP – OQ| or $\left|\vec{a}+\vec{b}\right|>\left|\left|\vec{a}\right|-\left|\vec{b}\right|\right|\dots$ (iii) In case, the vectors $\vec{a}$ and $\vec{b}$ the same straight line but point in the opposite direction, then $\left|\vec{a}+\vec{b}\right|=\left|\left|\vec{a}\right|-\left|\vec{b}\right|\right|\dots$ (iv) Combining the conditions stated in the equations (iii) and (iv), we have $\left|\vec{a}+\vec{b}\right|\ge\left|\left|\vec{a}\right|-\left|\vec{b}\right|\right|$ (c) To prove $\left|\vec{a}-\vec{b}\right|\le\left|\vec{a}\right|+\left|\vec{b}\right|$ In figure (ii) the vectors $\vec{a}$ and $\vec{b}$ are represented by $\overline{OL}$ and $\overline{OM}$ respectively. Therefore, the vector $\vec{a}-\vec{b}$ is given by $\overline{ON}$ From the ΔOMN, it follows that ON < MN + OM or $\left|\vec{a}-\vec{b}\right| < \left|\vec{a}\right|+\left|\vec{-b}\right|$ or $\left|\vec{a}-\vec{b}\right| < \left|\vec{a}\right|+\left|\vec{b}\right|\dots$ (v) In case, the vectors $\vec{a}$ and $\vec{b}$ are along the same straight line but point inthe opposite direction, then $\left|\vec{a}-\vec{b}\right| < \left|\vec{a}\right|+\left|\vec{b}\right|\dots$(vi) Combining the conditions stated in the equations (v) and (vi), we have $\left|\vec{a}-\vec{b}\right|\le\left|\vec{a}\right|+\left|\vec{b}\right|$ (d) To prove $\left|\vec{a}-\vec{b}\right|\ge\left|\left|\vec{a}\right|-\left|\vec{b}\right|\right|$ In figure (ii) again consider the ΔOMN. It follows that ON + OM > MN or ON > |MN – OM| The modulus of MN – OM has been taken for the reason that whereasL.H.S. is positive, R.H.S. may be negative, in case MN is smaller than OM. Since MN = OL, we have ON > |OL – OM| $\left|\vec{a}-\vec{b}\right|>\left|\left|\vec{a}\right|-\left|\vec{-b}\right|\right|$ or $\left|\vec{a}-\vec{b}\right|>\left|\left|\vec{a}\right|-\left|\vec{b}\right|\right|\dots$ (vii) In case, the vectors $\vec{a}$ and $\vec{b}$ are along the same straight line and point inthe same direction, then $\left|\vec{a}-\vec{b}\right|=\left|\left|\vec{a}\right|-\left|\vec{b}\right|\right|\dots$ (viii) Combining the conditions stated in equations (vii) and (viii), we have $\left|\vec{a}-\vec{b}\right|>\left|\left|\vec{a}\right|-\left|\vec{b}\right|\right|$