Question

Vector
a⃗
a→
of length 8 m lies at an angle of
60o
60o
above the x-axis in the first quadrant. Vector
b⃗
b→
of length 5 m lies
53o
53o
below the x-axis in the fourth quadrant. Determine the magnitude of
a⃗ −b⃗

Answer 1

Answer on Question #58021 - Physics - Mechanics - Relativity

$\left|\vec{a}\right| = 8m,\angle (Ox,\vec{a}) = 60{}^{\circ}$

$\left|\vec{b}\right| = 5m,\angle (Ox,\vec{b}) = -53{}^{\circ}$

For convenience denote:

$\left|\vec{a}\right| = l_{1},\angle (Ox,\vec{a}) = \gamma_{1}$

$\left|\vec{b}\right| = l_{2},\angle (Ox,\vec{b}) = \gamma_{2}$

Transfer vectors at the beginning of the coordinate system:

Coordinates of vectors:

$$\vec {a} = \left(x _ {1}, y _ {1}\right)$$ $$\vec {b} = (x _ {2}, y _ {2})$$

Search these coordinates:

$$y _ {1} = l _ {1} \sin \gamma_ {1}$$ $$x _ {1} = l _ {1} \cos \gamma_ {1}$$ $$y _ {2} = l _ {2} \sin \gamma_ {2}$$ $$x _ {2} = l _ {2} \cos \gamma_ {2}$$

Search difference of vectors:

$$\begin{array}{l} \vec {a} - \vec {b} = \overline {{(l _ {1} \cos \gamma_ {1} - l _ {2} \cos \gamma_ {2} , l _ {1} \sin \gamma_ {1} - l _ {2} \sin \gamma_ {2})}} = \\ = \overline {{(8 * c o s 6 0 {}^ {\circ} - 5 * c o s (- 5 3 {}^ {\circ}) , 8 * s i n 6 0 {}^ {\circ} - 5 * s i n (- 5 3 {}^ {\circ}))}} \approx \overline {{(0 , 9 9 , 1 0 , 9 2)}} \\ \end{array}$$

The magnitude:

$$\left| \vec {a} - \vec {b} \right| = \sqrt {\left(l _ {1} \cos \gamma_ {1} - l _ {2} \cos \gamma_ {2}\right) ^ {2} + \left(l _ {1} \sin \gamma_ {1} - l _ {2} \sin \gamma_ {2}\right) ^ {2}} \approx 1 0, 9 6 \mathrm {m}$$

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