Question

If $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ and $|\vec{a}| = 3, |\vec{b}| = 5, |\vec{c}| = 7$, then the angle between $\vec{a}$ and $\vec{b}$ is

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{4}$
• D. $\frac{\pi}{6}$

Hint:

This is a classic problem. When given a vector sum equal to zero, to find the angle between any two vectors (say A and B), isolate the third vector (C) on one side of the equation and then square the magnitudes of both sides. This introduces the dot product A.B, which contains the angle information.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We are given a relationship between three vectors and their magnitudes. To find the angle between two of the vectors, $\vec{a}$ and $\vec{b}$, we can use the dot product formula, which relates the dot product to the magnitudes and the cosine of the angle.
Step 2: Key Formula or Approach:
The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$. We can find the value of $\vec{a} \cdot \vec{b}$ by manipulating the given vector equation.
Step 3: Detailed Explanation:
We are given $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.
To find the angle between $\vec{a}$ and $\vec{b}$, we should isolate the third vector, $\vec{c}$.
\[ \vec{a} + \vec{b} = -\vec{c} \] Now, take the dot product of each side with itself. This is equivalent to squaring the magnitude of each side.
\[ (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (-\vec{c}) \cdot (-\vec{c}) \] \[ |\vec{a} + \vec{b}|^2 = |-\vec{c}|^2 = |\vec{c}|^2 \] Expand the left side:
\[ \vec{a} \cdot \vec{a} + 2(\vec{a} \cdot \vec{b}) + \vec{b} \cdot \vec{b} = |\vec{c}|^2 \] \[ |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = |\vec{c}|^2 \] Now, substitute the given magnitudes: $|\vec{a}| = 3, |\vec{b}| = 5, |\vec{c}| = 7$.
\[ 3^2 + 2(\vec{a} \cdot \vec{b}) + 5^2 = 7^2 \] \[ 9 + 2(\vec{a} \cdot \vec{b}) + 25 = 49 \] \[ 34 + 2(\vec{a} \cdot \vec{b}) = 49 \] \[ 2(\vec{a} \cdot \vec{b}) = 49 - 34 = 15 \] \[ \vec{a} \cdot \vec{b} = \frac{15}{2} \] Now we can find the angle $\theta$ between $\vec{a}$ and $\vec{b}$.
\[ \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{15/2}{3 \times 5} = \frac{15/2}{15} = \frac{1}{2} \] The angle $\theta$ in the range $[0, \pi]$ for which $\cos\theta = \frac{1}{2}$ is $\theta = \frac{\pi}{3}$.
Step 4: Final Answer:
The angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi{3}$}.