Question

please help me in this question
i will mark you brainliest

Answer 1

Your answer is --

Given,

|A+B| = A = B .......(1)

So,

magnitude of |A+B| is

√A^2+B^2+2ABcos@ = A (since |A+B| = A )

=> A^2+A^2+2A^2cos@ = A^2 (from 1)

=> 2A^2+2A^2cos@ = A^2

=> 2A^2(1+cos@) = A^2

=> 1+cos@ = 1/2

=> 1+cos@ = 1/2-1

=> cos@ = -1/2

=> @ = 120° [ cos120° = -1/2 ]


Now, magnitude of |A-B| is --

√A^2+B^2-2ABcos120°

= √A^2+A^2-2A^2cos120° ( from 1)

= √2A^2 - 2A^2×-1/2 (cos120° = -1/2)

= √2A^2+A^2

= √3A^2

= √3 |A|


Therefore , |A-B| = √3|A|


Hence, option (4) is right


【 Hope it helps you 】

Answer 2

i think the 3rd option will be the correct one