Question

Two vectors A and B have magnitude 6 units and 8 units respectively. Find |A-B|, If the angle between two vectors is 0°, 180°,60°,120°,90°​

Answer 1

Answer :  formulae is - |A - B| = √(A^2 + B^2 - 2ABcosθ)

Explanation : please follow this -

example 1

According to your magnitude of vectors there should be a condition which says that angle between them is 90 degree .

you have to use the formula of addition of vector

if vector A +vector B =vector C  

then |C|=(A^2 +B^2 + 2 AB cos x )^1/2

where x =90 ° /0/180/60/120 degree = thetha

example - 2

A = |A|

B = |B|

 

|A + B| = √(A2 + B2 + 2ABcosθ)

 

|A - B| = √(A2 + B2 - 2ABcosθ)

 

|A + B| = |A - B| ⇒

 

√(A2 + B2 + 2ABcosθ) = √(A2 + B2 - 2ABcosθ)

 

Square both sides.

 

A2 + B2 + 2ABcosθ = A2 + B2 - 2ABcosθ

 

2ABcosθ = -2ABcosθ

 

If A ≠ 0 and B ≠ 0, then

 

cosθ = -cosθ ⇒

 

cosθ = 0 ⇒ θ = 90°

in the example 2 , take a look from bottom to top ,hope you will understand