Question

What is the dot product of two vector of magnitude 3 and 5, if angle between them is 60 degree?

Answer 1

Answer:

See below.

Explanation:

Let the vectors be a and b. Let theta be the angle between them.

$|\vec{a}| = 3\\|\vec{b}| = 5\\\theta = 60^\circ\\\\\vec{a} . \vec{b} = |\vec{a}| |\vec{b}| \cos{60^\circ}\\= 15 \times \frac{1}{2}\\= 7.5$

Answer 2

Answer:

A.B=7.5

Explanation:

Given that

Vector A have magnitude= 3

Vector B have magnitude= 5

Angle between them θ =60 °

We know that

$cos\theta=\dfrac{A.B}{|A||B|}$

A.B is the dot product of two vector A and B

Given

|A|=3

|B|=5

θ =60 °

Now by putting the values

$cos\60^{\circ}=\dfrac{A.B}{3\times 5}$

$\dfrac{1}{2}=\dfrac{A.B}{3\times 5}$

A.B=7.5

So the dot product of vector is 7.5