Question

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

• A. \(30^\circ\)
• B. \(45^\circ\)
• C. \(60^\circ\)
• D. \(90^\circ\)

Hint:

If \(\vec{a}+\vec{b}+\vec{c}=0\), then one vector can be expressed as the negative of the sum of the other two vectors, and the magnitude formula can be used to find the angle between them.

Answer 1

The Correct Option is C

Step 1: Use the closing condition.
Given $\vec a+\vec b+\vec c=0$, so $\vec a=-(\vec b+\vec c)$.
Step 2: Take magnitudes squared.
Then $|\vec a|^2=|\vec b+\vec c|^2$.
Step 3: Expand the right side.
Let $\theta$ be the angle between $\vec b$ and $\vec c$. Then \[ |\vec b+\vec c|^2=|\vec b|^2+|\vec c|^2+2|\vec b||\vec c|\cos\theta. \] Step 4: Substitute the magnitudes.
With $|\vec a|=7$, $|\vec b|=5$, $|\vec c|=3$: \[ 49=25+9+2(5)(3)\cos\theta. \] Step 5: Solve for $\cos\theta$.
So $49=34+30\cos\theta$, giving $30\cos\theta=15$, hence $\cos\theta=\dfrac{1}{2}$.
Step 6: Read off the angle.
Therefore $\theta=60^\circ$.
\[ \boxed{60^\circ} \]