# Question #40a2f

Mar 24, 2017

(a) ${0}^{\circ}$
Since the magnitude of resultant is equal to the sum of magnitudes of two displacements.
$| \vec{R} | = | \vec{A} | + | \vec{B} |$

(b) ${180}^{\circ}$
Since the magnitude of resultant is equal to the difference of magnitudes of two displacements. (assuming $| \vec{A} | > | \vec{B} |$)
$| \vec{R} | = | \vec{A} | - | \vec{B} |$
(b) ${90}^{\circ}$
Since the square of magnitude of resultant is equal to sum of squares of magnitudes of two displacements. This is true only in case of a right triangle.
$| \vec{R} {|}^{2} = | \vec{A} {|}^{2} + | \vec{B} {|}^{2}$