Test Index

Motion in a Plane

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Question : 6
Total: 32
Establish the following vector inequalities geometrically or otherwise :
(a) |
→
a
+
→
b
|≤|
→
a
|+|
→
b
|
(b) |
→
a
+
→
b
|≥||
→
a
|+|
→
b
||
(c) |
→
a
−
→
b
|≤|
→
a
|+|
→
b
|
(b) | a− b|≥||
→
a
|−|
→
b
||
When does the equality sign above apply?
Solution:  
Consider that the two vectors
→
a
 and
→
b
 are represented by
––––
OP
 and
––––
OQ
 . The addition of the two vectors i.e.
→
a
+
→
b
 is given by
––––
OR
 asshown in figure (i)
(a) To prove |
→
a
+
→
b
|≤|
→
a
|+|
→
b
|
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=|
→
a
+
→
b
|,OP=|
→
a
| and OQ=|
→
b
|
∴ |
→
a
+
→
b
|<|
→
a
|+|
→
b
|... (i)
In case, the vectors
→
a
 and
→
b
 are along the same straight line and point inthe same direction, then
|
→
a
+
→
b
|=|
→
a
|+|
→
b
|... (ii)
Combining the conditions stated in the equations (i) and (ii), we have
|
→
a
+
→
b
|≤|
→
a
|+|
→
b
|
(b) To prove |
→
a
+
→
b
|≥||
→
a
|−|
→
b
||
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 |
→
a
+
→
b
|>||
→
a
|−|
→
b
||... (iii)
In case, the vectors
→
a
 and
→
b
 the same straight line but point in the opposite direction, then
|
→
a
+
→
b
|=||
→
a
|−|
→
b
||... (iv)
Combining the conditions stated in the equations (iii) and (iv), we have
|
→
a
+
→
b
|≥||
→
a
|−|
→
b
||
(c) To prove |
→
a
−
→
b
|≤|
→
a
|+|
→
b
|
In figure (ii) the vectors
→
a
 and
→
b
 are represented by
–––
OL
 and
–––
OM
 respectively. Therefore, the vector
→
a
−
→
b
 is given by
→
ON
From the ΔOMN, it follows that
ON < MN + OM
or |
→
a
−
→
b
| <|
→
a
|+|
→
−b
|
or |
→
a
−
→
b
| <|
→
a
|+|
→
b
|... (v)
In case, the vectors
→
a
 and
→
b
 are along the same straight line but point inthe opposite direction, then
|
→
a
−
→
b
| <|
→
a
|+|
→
b
|...(vi)
Combining the conditions stated in the equations (v) and (vi), we have
|
→
a
−
→
b
| ≤|
→
a
|+|
→
b
|
(d) To prove |
→
a
−
→
b
|≥||
→
a
|−|
→
b
||
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|
|
→
a
−
→
b
| >||
→
a
|−|
→
−b
||
or |
→
a
−
→
b
| >||
→
a
|−|
→
b
||... (vii)
In case, the vectors
→
a
 and
→
b
 are along the same straight line and point inthe same direction, then
|
→
a
−
→
b
| =||
→
a
|−|
→
b
||... (viii)
Combining the conditions stated in equations (vii) and (viii), we have
|
→
a
−
→
b
| >||
→
a
|−|
→
b
||
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