If a and b are two vectors such that | a + b | = | a |, then prove that vector 2 a + b is perpendicular to vector b .

| a + b | = | a | | a + b |^ 2 =| a |^ 2 ( a + b ).( a + b ) = | a |^ 2 a . a + a . b + b . a + b . b = | a |^ 2 | a |^ 2 + 2 a . b + b . b = | a |^ 2 [ a . b = b . a ] 2 a . b + b . b = 0 (2 a + b ). b = 0 ( 2 a + b ) is perpendicular to b .

If a and b are two vectors such that | a + b | =

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Question

If $\overrightarrow{\text{a}}\text{ and } \overrightarrow{\text{b}}$are two vectors such that $|\overrightarrow{\text{a}} + \overrightarrow{\text{b}}| = | \overrightarrow{\text{a}}|,$ then prove that vector 2 $\overrightarrow{\text{a}} + \overrightarrow{\text{b}}$is perpendicular to vector$\overrightarrow{\text{b}}.$

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Answer

$\because|\overrightarrow{\text{a}} + \overrightarrow{\text{b}}| = |\overrightarrow{\text{a}}|\Rightarrow|\overrightarrow{\text{a}} +\overrightarrow{\text{b}}|^{2} =|\overrightarrow{\text{a}}|^{2}$
$\Rightarrow(\overrightarrow{\text{a}} + \overrightarrow{\text{b}}).(\overrightarrow{\text{a}} + \overrightarrow{\text{b}}) = |\overrightarrow{\text{a}}|^{2}$
$\Rightarrow\overrightarrow{\text{a}}.\overrightarrow{\text{a}} + \overrightarrow{\text{a}}.\overrightarrow{\text{b}} + \overrightarrow{\text{b}}.\overrightarrow{\text{a}}+\overrightarrow{\text{b}}.\overrightarrow{\text{b}} = |\overrightarrow{\text{a}}|^{2}$
$\Rightarrow|\overrightarrow{\text{a}}|^{2} + 2 \overrightarrow{\text{a}}.\overrightarrow{\text{b}} +\overrightarrow{\text{b}}.\overrightarrow{\text{b}} = | \overrightarrow{\text{a}}|^{2}\big[\because\overrightarrow{\text{a}}.\overrightarrow{\text{b}} = \overrightarrow{\text{b}}.\overrightarrow{\text{a}}\big]$
$\Rightarrow2\overrightarrow{\text{a}}.\overrightarrow{\text{b}} + \overrightarrow{\text{b}}.\overrightarrow{\text{b}} = 0 \Rightarrow(2\overrightarrow{\text{a}} +\overrightarrow{\text{b}}).\overrightarrow{\text{b}} = 0 $
$\Rightarrow( 2 \overrightarrow{\text{a}} + \overrightarrow{\text{b}})$ is perpendicular to $\overrightarrow{\text{b}}.$

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