Question

CHAPTER : Motion in a Plane .

If for two vectors vector A and vector B , sum of A + B is perpendicular to difference A - B , the ratio of magnitudes is :
(1) 1:1
(2) 2:1
(3) 3:1
(4) 4:1

Answer 1

$\vec{A}+\vec{B}$ and $\vec{A}-\vec{B}$ are perpendicular to each other .

By scalar product of vector multiplication we have :

$\vec{X}.\vec{Y}=|X||Y|cos\theta$ where $\theta$ is the angle between the two vectors X and Y .

Given :

$(\vec{A}+\vec{B}).(\vec{A}-\vec{B})=|A+B||A-B|cos\theta\\\\\implies (\vec{A}+\vec{B}).(\vec{A}-\vec{B})=|A+B||A-B|cos90^\circ\\\\\implies (\vec{A}+\vec{B}).(\vec{A}-\vec{B})=0\\\\\therefore\mathsf{cos90^\circ=0}\\\\\implies \vec{A}.\vec{A}+\vec{A}.\vec{B}-\vec{A}.\vec{B}-\vec{B}\vec{B}=0\\\\\implies A^2cos0^\circ-B^2cos0^\circ=0\\\\\implies A^2-B^2=0\\\\\implies A^2=B^2$

$A=\pm B$

Neglect the negative value because magnitude cannot be negative .

$A:B=1:1$

Option (1) is correct .