Question

If A and B are perpendicular to each other, prove that IA +B I = √A2 + B2

Answer 1

Answer:

Proven

Explanation:

Squaring the LHS :-

|A+B|^2 = |A|^2 + |B|^2 + 2|AB|

= |A|^2 + |B|^2 + 0 because the dot product of two orthogonal vectors is 0.

Therefore, |A+B|^2 = |A|^2 + |B|^2

so, taking the square root on both sides

|A+B| = sqrt(|A|^2 + |B|^2)

hence proved !

Answer 2

To make a proof that $(a+b)^2 = (a-b)^2$

Given:

The vectors  A and B are perpendicular to each other

To Find:

To prove that  IA +B I = √A2 + B2

Solution:

Let a and b are perpendicular to each other,

a.b = b.a = 0......(1)

Considering the LHS,

= ($a+b)^2$

= (a+b)(a+b)

= a(a+b) + b(a+b)

= a.a +0 +0+b.b.....by(1)

= $|a|^2 +|b|^2$

Consider, the RHS Part

We get,

= (a-b)^2

= (a-b)(a-b)

= a.(a-b) + b(a-b)

= a.a +b.b

=  $|a|^2 +|b|^2$

LHS = RHS

Hence, we get $(a+b)^2 = (a-b)^2$

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