Question

If the magnitude of sum of two non zero vectors is equal to the magnitude of their difference then which of the following is true?
a Two vectors must have equal magnitude
b Angle between two vectors must be 90°
c Two vectors have equal magnitude but oppositely directed
d Angle between the vectors must be, 120°​

Answer 1

Given:

Magnitude of sum of two non-zero vectors= the magnitude of their difference

i.e. |a→ + b→| = |a→ - b→|

To find:

Which of the given options are correct.

Solution:

Since the magnitude of the sum of two non-zero vectors is equal to the magnitude of their difference.

Let the vectors be a→ and b→.

|a→ + b→| = |a→ - b→|

a²+b² - 2abcosα = a²+b² + 2ab cosα

4ab cosα = 0

cos α = 0

α = 90°

Since the vectors are perpendicular, the correct option is b.

The angle between two vectors must be 90° is the correct option.

Answer 2

SOLUTION :-

GIVEN :-

The magnitude of sum of two non zero vectors is equal to the magnitude of their difference

TO CHOOSE THE CORRECT OPTION :-

a. Two vectors must have equal magnitude

b. Angle between two vectors must be 90°

c. Two vectors have equal magnitude but oppositely directed

d. Angle between the vectors must be, 120°

FORMULA TO BE IMPLEMENTED :-

$1. \: \sf{For \: any \: two \: vectors\: \: \vec{a} \: \: and \: \: \vec{b} \: }$

$\sf{ \vec{a} . \: \vec{b} = | \vec{a}| | \vec{b}| \cos \theta }$

$\sf{ where \: \: \theta \: \: is \: the \: angle\:between \: \vec{a} \: \: and \: \: \vec{b} }$

$\sf{2. \: \: \vec{a}. \vec{a} = \: { | \vec{a} |}^{2}}$

EVALUATION :-

$\sf{ Let\: \: \vec{a} \: \: and \: \: \vec{b} \: \:are \: two \: non \: zero \: vectors }$

$\sf{ \: and \: \: \theta \: \: is \: the \: angle\:between \: \vec{a} \: \: and \: \: \vec{b} }$

Here it is given that

$\sf{ | \: \vec{a} + \vec{b} \: | = | \: \vec{a} - \vec{b} \: | \: }$

Squaring both sides we get

$\sf{ {| \: \vec{a} + \vec{b} \: |}^{2} = {| \: \vec{a} - \vec{b} \: |}^{2} \: }$

$\implies\sf{ ( \vec{a} + \vec{b} ). ( \vec{a} + \vec{b} ) = ( \vec{a} - \vec{b} ). ( \vec{a} - \vec{b} )} \: ( \: by \: formula \: 2)$

$\implies \sf{ \vec{a} . \vec{a} + \vec{a} .\vec{b} +\vec{b} .\vec{a} + \vec{b} .\vec{b} = \vec{a} . \vec{a} - \vec{a} .\vec{b} - \vec{b} .\vec{a} + \vec{b} .\vec{b} \: }$

$\implies \sf{ {| \vec{a}|}^{2} + 2\vec{a} .\vec{b} +{ | \vec{b}|}^{2} = {| \vec{a}|}^{2} - 2\vec{a} .\vec{b} +{ | \vec{b}|}^{2} } \: ( \because \: vector \: dot \: product \: is \: commutative \: )$

$\implies \sf{ 4 \: \vec{a} .\vec{b} = 0 }$

$\implies \sf{ \vec{a} .\vec{b} = 0 }$

$\implies \sf{ | \vec{a}| | \vec{b}| \cos \theta} = 0 \: \: (by \: formula \: 1)$

$\implies \sf{ \cos \theta} = 0 \: (\because \: | \vec{a}| \ne 0 \: and \: | \vec{b}| \ne 0 )$

$\implies \sf{ \theta = {90}^{ \circ} }$

FINAL ANSWER :-

The correct option is

b. Angle between two vectors must be 90°

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