Question

If \( \vec{a}, \vec{b}, \vec{c} \) are vectors such that \( \vec{a} + \vec{b} + \vec{c} = 0 \) and \( |\vec{a}| = 7, |\vec{b}| = 5, |\vec{c}| = 3 \), then the angle between \( \vec{c} \) and \( \vec{b} \) is:

• A. \( \frac{\pi}{3} \)
• B. \( \frac{\pi}{6} \)
• C. \( \frac{\pi}{4} \)
• D. \( \pi \)
• E. \( 0 \)

Hint:

This is effectively the Law of Cosines for the triangle formed by these vectors. Always isolate the vector that does NOT bound the angle you are trying to find.

Answer 1

The Correct Option is A

Concept: If the sum of three vectors is zero, they form a triangle. The relationship between their magnitudes and the angle \( \theta \) between any two vectors can be found by isolating the third vector and squaring both sides of the equation.

Step 1: Isolate \( \vec{a} \) and square the equation.
Given \( \vec{a} + \vec{b} + \vec{c} = 0 \), then: \[ \vec{b} + \vec{c} = -\vec{a} \] Square both sides (taking the dot product with itself): \[ (\vec{b} + \vec{c}) \cdot (\vec{b} + \vec{c}) = (-\vec{a}) \cdot (-\vec{a}) \] \[ |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{b} \cdot \vec{c}) = |\vec{a}|^2 \] \[ |\vec{b}|^2 + |\vec{c}|^2 + 2|\vec{b}||\vec{c}|\cos \theta = |\vec{a}|^2 \]

Step 2: Substitute magnitudes and solve for \( \cos \theta \).
\[ 5^2 + 3^2 + 2(5)(3)\cos \theta = 7^2 \] \[ 25 + 9 + 30\cos \theta = 49 \] \[ 34 + 30\cos \theta = 49 \] \[ 30\cos \theta = 15 \] \[ \cos \theta = \frac{15}{30} = \frac{1}{2} \]

Step 3: Find the angle.
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \]