Question
Establish the following vector inequalities geometrically or otherwise: $|\text{a}-\text{b}|\le|\text{a}|+|\text{b}|$ When does the equality sign above apply?
✓
Answer
Let two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.
$|\vec{\text{OR}}|=|\vec{\text{PS}}|=|\vec{\text{b}}|\ ...(\text{i})$ $|\vec{\text{OP}}|=|\vec{\text{a}}|\ ...(\text{ii})$ In a triangle, each side is smaller than the sum of the other two sides. Therefore, in $\triangle\text{OPS},$ $\text{OS}<\text{OP}+\text{PS}$ $|\vec{\text{a}}-\vec{\text{b}}|<|\vec{\text{a}}|+|-\vec{\text{b}}|$ $|\vec{\text{a}}-\vec{\text{b}}|<|\vec{\text{a}}|+|\vec{\text{b}}|\ ...(\text{iii})$ If the two vectors act in a straight line but in opposite directions, then we can write: $|\vec{\text{a}}-\vec{\text{b}}|=|\vec{\text{a}}|+|\vec{\text{b}}|\ ...(\text{iv})$ Combining equations (iii) and (iv), we get: $|\vec{\text{a}}-\vec{\text{b}}|\le|\vec{\text{a}}|+|\vec{\text{b}}|$
$|\vec{\text{OR}}|=|\vec{\text{PS}}|=|\vec{\text{b}}|\ ...(\text{i})$ $|\vec{\text{OP}}|=|\vec{\text{a}}|\ ...(\text{ii})$ In a triangle, each side is smaller than the sum of the other two sides. Therefore, in $\triangle\text{OPS},$ $\text{OS}<\text{OP}+\text{PS}$ $|\vec{\text{a}}-\vec{\text{b}}|<|\vec{\text{a}}|+|-\vec{\text{b}}|$ $|\vec{\text{a}}-\vec{\text{b}}|<|\vec{\text{a}}|+|\vec{\text{b}}|\ ...(\text{iii})$ If the two vectors act in a straight line but in opposite directions, then we can write: $|\vec{\text{a}}-\vec{\text{b}}|=|\vec{\text{a}}|+|\vec{\text{b}}|\ ...(\text{iv})$ Combining equations (iii) and (iv), we get: $|\vec{\text{a}}-\vec{\text{b}}|\le|\vec{\text{a}}|+|\vec{\text{b}}|$
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