Prove that ( a + b) ( a + b) = | a | 2 + | b | 2 , if and only if a, b are perpendicular, given a 0, b 0 .

( a + b) ( a + b) = a ( a + b) + b ( a + b) = a a + a b + b a + b b = | a | 2 + a b + b a + | b | 2 = | a | 2 + | b | 2 + 2 a b …(i) When a, b are perpendicular a b = 0 From (i), ( a + b) ( a + b) = | a | 2 + | b | 2 Conversely, ( a + b) ( a + b) = | a | 2 + | b | 2 a b = 0 a, b are perpendicular.

Vector Algebra — Class 12 Maths Solution

ncert misc SA NCERT,Page 459,Misc,Q.No 15

Question

Prove that $(\vec a + \vec b) \cdot (\vec a + \vec b) = |\vec a{|^2} + |\vec b{|^2}$, if and only if $\vec a,\vec b$ are perpendicular, given $\vec a \ne \vec 0,\vec b \ne \vec 0$.

Step-by-step Solution

$(\vec a + \vec b) \cdot (\vec a + \vec b) = \vec a \cdot (\vec a + \vec b) + \vec b \cdot (\vec a + \vec b)$

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

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

When $\vec a,\vec b$ are perpendicular $\Rightarrow \vec a \cdot \vec b = 0$
$\therefore$

From (i), $(\vec a + \vec b) \cdot (\vec a + \vec b) = |\vec a{|^2} + |\vec b{|^2}$

Conversely, $(\vec a + \vec b) \cdot (\vec a + \vec b) = |\vec a{|^2} + |\vec b{|^2}$
$\Rightarrow \vec a \cdot \vec b = 0 \Rightarrow \vec a,\vec b$ are perpendicular.

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