Question

If $\vec{a}$ and $\vec{b}$ are unit vectors, such that $\vec{a} + 3\vec{b}$ is perpendicular to $7\vec{a} - 5\vec{b}$. The angle between $\vec{a}$ and $\vec{b}$ is

• A. 30 degrees
• B. 60 degrees
• C. 90 degrees
• D. 120 degrees

Answer 1

Answer: (B)

60 degrees

Answer 2

Given that $\vec{a}$ and $\vec{b}$ are unit vectors, we have $|\vec{a}| = 1$ and $|\vec{b}| = 1$. The condition that $(\vec{a} + 3\vec{b})$ is perpendicular to $(7\vec{a} - 5\vec{b})$ means their dot product is zero: $(\vec{a} + 3\vec{b}) \cdot (7\vec{a} - 5\vec{b}) = 0$ Expanding the dot product: $7(\vec{a} \cdot \vec{a}) - 5(\vec{a} \cdot \vec{b}) + 21(\vec{b} \cdot \vec{a}) - 15(\vec{b} \cdot \vec{b}) = 0$ Using $\vec{a} \cdot \vec{a} = |\vec{a}|^2$, $\vec{b} \cdot \vec{b} = |\vec{b}|^2$, and $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$: $7|\vec{a}|^2 + 16(\vec{a} \cdot \vec{b}) - 15|\vec{b}|^2 = 0$ Substitute $|\vec{a}| = 1$ and $|\vec{b}| = 1$: $7(1)^2 + 16(\vec{a} \cdot \vec{b}) - 15(1)^2 = 0$ $7 + 16(\vec{a} \cdot \vec{b}) - 15 = 0$ $16(\vec{a} \cdot \vec{b}) - 8 = 0$ $16(\vec{a} \cdot \vec{b}) = 8$ $\vec{a} \cdot \vec{b} = \frac{8}{16} = \frac{1}{2}$ The dot product is also defined as $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$. $\frac{1}{2} = (1)(1) \cos \theta$ $\cos \theta = \frac{1}{2}$ The angle $\theta$ for which $\cos \theta = \frac{1}{2}$ is $60^\circ$.