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Section Formula

For any two v...

Question

For any two vectors $^a$ and $^b$ prove that
(a) $| \to a + \to b | \leq | \to a | + | \to b |$
(b) $| \to a - \to b | \leq | \to a | + | \to b |$

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Solution

$| \to a + \to b | \leq | \to a | + | \to b |$
$(a)$ Let two vectors $\to a$ and $\to b$ represent the adjacent sides of a parallelogram $O M N P$ , as shown in the given figure.
Here we can write,
$O H = | a | - - - - (i)$
$M N = O P = | b | - - - - (i i)$
$O N = | \to a + \to b | - - - - (i i i)$
In a triangle each side is smaller than the sum of others two sides
Therefore, In $△ O M N$ , we have
$O N < (O M + M N)$
$| \to a + \to b | < | \to a | + | \to b | - - - - (i v)$
If the vectors $\to a$ and $\to b$ act along a straight line in the same direction, then we can write:
$| \to a + \to b | = | \to a | + | \to b | - - - - (v)$
Combining $(i v)$ and $(v)$ , we get
$| \to a | + | \to b | \geq | \to a + \to b |$
Let the two vectors $\to a$ and $\to b$ represent adjacent sides of a parallelogram $P Q R S$ as shown in figure.
Here, we have
$| O R | = | P S | = | \to b | - - - (i)$
$| O P | = | \to a | - - (i i)$
In a triangle each sides is smaller thab the sum of the other two sides. Through in $△ O P S$ we have
$O S < O P + P S$
$| \to a - \to b | < | \to a | + | - \to b |$
$| \to a - \to b | < | \to a | + | \to b | - - - (i i i)$
of two vectors act in straight line but in opposite direction, then we can write
$| \to a - \to b | = | \to a | + | \to b | \to (i v)$
Combining equation $(i i i)$ and $(i v)$ . we get $| \to a - \to b | \leq | \to a | + | \to b |$

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