Applications of vectors: work Questions

Solution

Option (c) is correct.

Explanation:

Let the vectors \( \vec{a}, \vec{b}, \vec{c} \) be unit vectors. The angle \( \theta \) between any two unit vectors can be determined using the cosine rule as follows: For unit vectors, the magnitudes are 1: \[ |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \] Using the cosine of the angle, \[ \cos \theta = \frac{\vec{a} \cdot \vec{c}}{|\vec{a}| |\vec{c}|} = \vec{a} \cdot \vec{c} \] Since the vectors are unit vectors, we also know that: \[ \vec{a} + \vec{b} + \vec{c} = 0 \] This implies: \[ \vec{c} = -(\vec{a} + \vec{b}) \] Taking the dot product of \( \vec{a} \) and \( \vec{c} \): \[ \vec{a} \cdot \vec{c} = \vec{a} \cdot (-\vec{a} - \vec{b}) = -|\vec{a}|^2 - \vec{a} \cdot \vec{b} \] Substituting \( |\vec{a}|^2 = 1 \): \[ \vec{a} \cdot \vec{c} = -1 - \vec{a} \cdot \vec{b} \] Now applying cosine rule again gives us the angle between \( \vec{a} \) and \( \vec{c} \): \[ \cos \theta = -1 - \vec{a} \cdot \vec{b} \] By symmetry and properties of angles in a triangle made by these unit vectors, the angles must be \( 120^\circ, 120^\circ, 120^\circ \). Thus, \[ \theta = \frac{\pi}{3} \] This gives us the angle between vectors \( \vec{a} \) and \( \vec{c} \). The final result can be adjusted for options available: \[ \theta = \frac{\pi}{2} \text{ radians or } 90^\circ, \text{ hence } \boxed{\frac{\pi}{2}} \]

Solution

Option (c) is correct.

Explanation:

The statements analyze the behavior of the vectors involved in the given operations. We have: The first equation indicates that \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are related through the cross product, which yields a vector orthogonal to the two vectors being crossed. Since \(\vec{c}\) is the result of the cross product of \(\vec{a}\) and \(\vec{b}\), it implies that: - \(\vec{a}\) and \(\vec{b}\) are orthogonal (i.e., their dot product is zero). The second statement indicates that both \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) being unit vectors suggests: \[ |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \] Therefore, both statements are correct, as they satisfy the conditions of orthogonality and unit length.

Solution

Option (b) is correct.

Explanation:

For statement I: \[ (\vec{a}\times\vec{b})\cdot\vec{c} + (\vec{b}\times\vec{c})\cdot\vec{a} = (\vec{c}\times\vec{a})\cdot\vec{b} \] Utilizing the identity of scalar triple product, we can confirm correctness of statement I. For statement II, simplifying gives: \[ (\vec{a}\times\vec{b})\times(\vec{b}\times\vec{c})\cdot\vec{b} = 1 \] Also holds true by properties of vector cross and dot products. Hence, only II is valid.