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Q4.6(c):
Establish the following vector inequalities geometrically or otherwise : (c) $|a-b| < |a| + |b|$. When does the equality sign above apply?

Solution :

Initial Theoretical Foundation

To establish the vector inequality $| \vec{a} - \vec{b} | \leq | \vec{a} | + | \vec{b} |$, we utilize the properties of vector addition and the Triangle Inequality theorem in vector space. Consider two vectors $\vec{a}$ and $\vec{b}$. The vector subtraction $\vec{a} - \vec{b}$ is defined as $\vec{a} + (-\vec{b})$.

Step 1: Geometric Representation

In a triangle formed by vectors $\vec{a}$ and $-\vec{b}$ placed head-to-tail, the resultant vector is $\vec{R} = \vec{a} + (-\vec{b}) = \vec{a} - \vec{b}$. By the Triangle Inequality Theorem, the length of any one side of a triangle is strictly less than the sum of the lengths of the other two sides, unless the vectors are collinear.

|a| |-b| = |b| |a - b|

Step 2: Analytical Derivation via Magnitude Formula

The magnitude of the difference of two vectors is given by the law of cosines [Per the Parallelogram Law of Vector Addition]:

$| \vec{a} - \vec{b} | = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 - 2|\vec{a}||\vec{b}|\cos\theta}$

Where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

We compare this to the scalar sum $(|\vec{a}| + |\vec{b}|)$:

$(|\vec{a}| + |\vec{b}|)^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|$

Since $-2|\vec{a}||\vec{b}|\cos\theta \leq 2|\vec{a}||\vec{b}|$ (because $\cos\theta \geq -1$), it follows that:

$|\vec{a}|^2 + |\vec{b}|^2 - 2|\vec{a}||\vec{b}|\cos\theta \leq |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|$

Taking the square root of both sides, we obtain $| \vec{a} - \vec{b} | \leq | \vec{a} | + | \vec{b} |$.

Step 3: Determination of Equality Condition

The equality $| \vec{a} - \vec{b} | = | \vec{a} | + | \vec{b} |$ holds when the triangle collapses into a straight line. This occurs when the vectors $\vec{a}$ and $-\vec{b}$ are parallel and in the same direction.

  • The direction of $-\vec{b}$ is opposite to $\vec{b}$.
  • For $\vec{a}$ and $-\vec{b}$ to be in the same direction, $\vec{a}$ must be in the opposite direction of $\vec{b}$.
  • Thus, $\vec{a}$ and $\vec{b}$ must be anti-parallel (i.e., the angle between them $\theta = 180^{\circ}$).
Condition Resulting Inequality
General case ($\theta \neq 180^{\circ}$) $| \vec{a} - \vec{b} | < | \vec{a} | + | \vec{b} |$
Anti-parallel ($\theta = 180^{\circ}$) $| \vec{a} - \vec{b} | = | \vec{a} | + | \vec{b} |$

Final Solution: The inequality $| \vec{a} - \vec{b} | \leq | \vec{a} | + | \vec{b} |$ is a direct consequence of the triangle inequality. The equality sign applies if and only if the vectors $\vec{a}$ and $\vec{b}$ are anti-parallel to each other.


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