Vector Sum and Angle Calculation

We are given a problem involving three vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$, with specific magnitudes and a relationship between them. The key information provided is:

• The vector sum: $\vec{a} + \vec{b} + \vec{c} = \vec{0}$
• The magnitudes of the vectors: $|\vec{a}| = 3$, $|\vec{b}| = 5$, $|\vec{c}| = 7$

Our objective is to determine the angle between vectors $\vec{a}$ and $\vec{b}$.

Relating Vectors Using the Given Sum

The fundamental equation that connects the vectors is $\vec{a} + \vec{b} + \vec{c} = \vec{0}$. To find the angle between $\vec{a}$ and $\vec{b}$, we need to utilize the dot product, which inherently involves these two vectors. Let's rearrange the equation to isolate the terms containing $\vec{a}$ and $\vec{b}$:

$ \vec{a} + \vec{b} = -\vec{c} $

Applying the Dot Product for Angle Calculation

The dot product is directly related to the angle between vectors. We can compute the dot product of both sides of the equation $\vec{a} + \vec{b} = -\vec{c}$ with itself:

$ (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (-\vec{c}) \cdot (-\vec{c}) $

Expanding the left side of the equation, we get:

$ \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} = \vec{c} \cdot \vec{c} $

Using the properties of the dot product, specifically $\vec{v} \cdot \vec{v} = |\vec{v}|^2$ and the commutative property $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$, the equation simplifies to:

$ |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = |\vec{c}|^2 $

We know that the dot product $\vec{a} \cdot \vec{b}$ can be expressed as $|\vec{a}| |\vec{b}| \cos \theta$, where $\theta$ represents the angle between vectors $\vec{a}$ and $\vec{b}$. Substituting this definition:

$ |\vec{a}|^2 + 2 |\vec{a}| |\vec{b}| \cos \theta + |\vec{b}|^2 = |\vec{c}|^2 $

Substituting Magnitudes and Solving for the Angle

Now, we substitute the given magnitudes of the vectors into the derived equation:

$ (3)^2 + 2 (3) (5) \cos \theta + (5)^2 = (7)^2 $

Calculate the square of each magnitude:

$ 9 + 2 (3) (5) \cos \theta + 25 = 49 $

Simplify the terms involving multiplication:

$ 9 + 30 \cos \theta + 25 = 49 $

Combine the constant terms on the left side:

$ 34 + 30 \cos \theta = 49 $

To find $\cos \theta$, we first isolate the term containing it:

$ 30 \cos \theta = 49 - 34 $

$ 30 \cos \theta = 15 $

Now, solve for $\cos \theta$ by dividing both sides by 30:

$ \cos \theta = \frac{15}{30} $

$ \cos \theta = \frac{1}{2} $

To determine the angle $\theta$, we find the inverse cosine (arccos) of $\frac{1}{2}$:

$ \theta = \arccos\left(\frac{1}{2}\right) $

The angle whose cosine value is $\frac{1}{2}$ is $\frac{\pi}{3}$ radians.

Therefore, the angle between vectors $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{3}$.