Answer
(a) equal to
(b) $7.8cm\ at\ 153^{\circ},\ 7.8cm\ at\ 63^{\circ}$
Work Step by Step
(a) Since both vectors have equal length of components (that is, 3.5 cm and 7.0 cm), the magnitude of displacement 1 is equal to displacement 2.
(b) First of all, we find the components of the two vectors
$\vec{\Delta r_1}=(-7.0cm)\hat x+(3.5cm)\hat y$
$\vec{\Delta r_2}=(3.5cm)\hat x+(7.0cm)\hat y$
Now we can find the magnitude and direction of the first vector
$\Delta r_1=\sqrt{(-7.0cm)^2+(3.5cm)^2}=7.8cm$
$\theta_1=tan^{-1}(\frac{3.5cm}{-7.0cm})=-27^{\circ}+180^{\circ}=153^{\circ}$
and the magnitude and the direction of second vector is given as:
$\Delta r_2=\sqrt{(3.5cm)^2+(7.0cm)^2}=7.8cm$
$\theta_1=tan^{-1}(\frac{7.0cm}{3.5cm})=63^{\circ}$
