Question:medium

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

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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.
Updated On: Mar 27, 2026
  • $\frac{\pi}{2}$
  • $\frac{\pi}{3}$
  • $\frac{\pi}{4}$
  • $\frac{\pi}{6}$
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The Correct Option is B

Solution and Explanation

Step 1: Conceptualization:
The problem involves a relationship between three vectors and their magnitudes. The objective is to determine the angle between two specific vectors, $\vec{a}$ and $\vec{b}$, utilizing the dot product formula. This formula connects the dot product of two vectors to their magnitudes and the cosine of the angle between them.
Step 2: Core Principle:
The angle $\theta$ between vectors $\vec{a}$ and $\vec{b}$ is defined by the equation $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$. The value of $\vec{a} \cdot \vec{b}$ can be derived by algebraic manipulation of the given vector equation.
Step 3: Derivation Process:
Given the equation $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.
To find the angle between $\vec{a}$ and $\vec{b}$, isolate $\vec{c}$:
\[ \vec{a} + \vec{b} = -\vec{c} \]Taking the dot product of both sides with themselves is equivalent to squaring the magnitude of each side:
\[ (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (-\vec{c}) \cdot (-\vec{c}) \]\[ |\vec{a} + \vec{b}|^2 = |-\vec{c}|^2 = |\vec{c}|^2 \]Expanding the left side yields:
\[ \vec{a} \cdot \vec{a} + 2(\vec{a} \cdot \vec{b}) + \vec{b} \cdot \vec{b} = |\vec{c}|^2 \]\[ |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = |\vec{c}|^2 \]Substitute the given magnitudes: $|\vec{a}| = 3, |\vec{b}| = 5, |\vec{c}| = 7$.
\[ 3^2 + 2(\vec{a} \cdot \vec{b}) + 5^2 = 7^2 \]\[ 9 + 2(\vec{a} \cdot \vec{b}) + 25 = 49 \]\[ 34 + 2(\vec{a} \cdot \vec{b}) = 49 \]Solve for the dot product:
\[ 2(\vec{a} \cdot \vec{b}) = 49 - 34 = 15 \]\[ \vec{a} \cdot \vec{b} = \frac{15}{2} \]Now, calculate the angle $\theta$ between $\vec{a}$ and $\vec{b}$:
\[ \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{15/2}{3 \times 5} = \frac{15/2}{15} = \frac{1}{2} \]For $\theta$ in the range $[0, \pi]$, the angle satisfying $\cos\theta = \frac{1}{2}$ is $\theta = \frac{\pi}{3}$.
Step 4: Conclusion:
The angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{3}$.
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