Question

\( \vec{a}, \vec{b}, \vec{c} \) are three mutually perpendicular unit vectors. If \( \theta \) is the angle between \( \vec{a} \) and \( (2\vec{a} + 3\vec{b} + 6\vec{c}) \), find the value of \( \cos \theta \).

Hint:

To find the angle between a vector and a combination of vectors, use the dot product and magnitude properties, and simplify using orthogonality conditions if applicable.

Answer 1

Step 1: Given conditions
\[ |\vec{a}| = |\vec{b}| = |\vec{c}| = 1, \quad \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0. \] 
Step 2: Find the magnitude of \( 2\vec{a} + 3\vec{b} + 6\vec{c} \)
The magnitude is: \[ |2\vec{a} + 3\vec{b} + 6\vec{c}|^2 = (2^2)|\vec{a}|^2 + (3^2)|\vec{b}|^2 + (6^2)|\vec{c}|^2 = 4 + 9 + 36 = 49. \] Thus: \[ |2\vec{a} + 3\vec{b} + 6\vec{c}| = \sqrt{49} = 7. \] 
Step 3: Find \( \cos \theta \)
The dot product is: \[ \vec{a} \cdot (2\vec{a} + 3\vec{b} + 6\vec{c}) = (2\vec{a} \cdot \vec{a}) + (3\vec{a} \cdot \vec{b}) + (6\vec{a} \cdot \vec{c}). \] Since \( \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0 \), this simplifies to: \[ \vec{a} \cdot (2\vec{a} + 3\vec{b} + 6\vec{c}) = 2|\vec{a}|^2 = 2. \] Thus: \[ \cos \theta = \frac{\vec{a} \cdot (2\vec{a} + 3\vec{b} + 6\vec{c})}{|\vec{a}||2\vec{a} + 3\vec{b} + 6\vec{c}|} = \frac{2}{1 \cdot 7} = \frac{2}{7}. \] Conclusion: 
The value of \( \cos \theta \) is \( \frac{2}{7} \).