Question

Two vectors
a⃗
and
b⃗
have components, in arbitrary units,
ax=3.2
,
ay=1.6
,
bx=0.5
,
by=4.5
. Find the angle between
a⃗
and
b⃗

33o

28o

57o

62o

Answer 1

Answer on Question #51857-Physics-Mechanics-Kinematics-Dynamics

Two vectors $a$ and $b$ have components, in arbitrary units, $ax = 3.2, ay = 1.6, bx = 0.5, by = 4.5$ . Find the angle between $a$ and $b$ .

33o 28o 57o 62o

Solution

The scalar product of two vectors is

$$(\vec {a}, \vec {b}) = | \vec {a} | \cdot | \vec {b} | \cos \alpha = (a _ {x} b _ {x} + a _ {y} b _ {y}),$$

where $\alpha$ is the angle between $a$ and $b$ , $|\vec{a}|$ is the length of $\vec{a}$ , $|\vec{b}|$ is the length of $\vec{b}$ .

Thus,

$$\alpha = \cos^ {- 1} \left(\frac {a _ {x} b _ {x} + a _ {y} b _ {y}}{\sqrt {a _ {x} ^ {2} + a _ {y} ^ {2}} \sqrt {b _ {x} ^ {2} + b _ {y} ^ {2}}}\right) = \cos^ {- 1} \left(\frac {3 . 2 \cdot 0 . 5 + 1 . 6 \cdot 4 . 5}{\sqrt {3 . 2 ^ {2} + 1 . 6 ^ {2}} \sqrt {0 . 5 ^ {2} + 4 . 5 ^ {2}}}\right) = 5 7 {}^ {\circ}.$$

Answer: $57{}^{\circ}$

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