For any two vectors a and b - we always have - a - b - - a - - b - triangle inequality

| a #160; + b|^2 = | a|^2 + | b|^2 + 2| a|| b|.cosθ | a|^2 + | b|^2 = | a|^2 + | b|^2 + 2| a| + | b| ∵ #160; 1 ⩽cosθ #160; ⩽ 1 Righ ...

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Triangle Inequality

For any two v...

Question

For any two vectors $\to a$ and $\to b$ , we always have $∣ ∣ ∣ \to a + \to b ∣ ∣ ∣ \leq ∣ ∣ \to a ∣ ∣ + ∣ ∣ ∣ \to b ∣ ∣ ∣$ (triangle inequality)

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Establish the following vector inequalities geometrically or
otherwise:

(a) $| a + b | \leq | a | + | b |$

(b) $| a + b | \geq | | a | - | b | |$

| a \cdot b | \leq | a | \cdot | b |

b

c

a

(a \cdot b) b + (a \cdot c) c + \frac{a \cdot (b \times c)}{{| b \times c |}^{2}} (b \times c)

\to b

\to c

\to a

(\to a . \to b) \to b + (\to a . \to c) \to c + \frac{\to a (\to b \times \to c)}{{∣ ∣ ∣ \to b \times \to c ∣ ∣ ∣}^{2}} (\to b \times \to c)

b

c

a

(a . b) b + (a . c) + \frac{a . b \times c}{| b \times c |} . (b \times c)

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