Question

If \( \vec{A}, \vec{B} \) are unit vectors. If \( \vec{C} = \vec{A} + 2\vec{B} \) and \( \vec{D} = 5\vec{A} - 4\vec{B} \) are perpendicular, then angle between \( \vec{A} \) and \( \vec{B} \) is:

• A. \( \pi/2 \)
• B. \( \pi/3 \)
• C. \( \pi/4 \)
• D. None of the above

Hint:

Remember that two non-zero vectors are perpendicular if and only if their dot product is zero. Also, recall that \( \vec{A} \cdot \vec{A} = |\vec{A}|^2 \) and \( \vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta \). For unit vectors, \( |\vec{A}|=1 \) and \( |\vec{B}|=1 \).

Answer 1

The Correct Option is B

Step 1: Understand the given conditions. We are given that \( \vec{A} \) and \( \vec{B} \) are unit vectors, which means \( |\vec{A}| = 1 \) and \( |\vec{B}| = 1 \). We are also given two vectors \( \vec{C} = \vec{A} + 2\vec{B} \) and \( \vec{D} = 5\vec{A} - 4\vec{B} \). The condition is that \( \vec{C} \) and \( \vec{D} \) are perpendicular. Step 2: Use the condition for perpendicular vectors. Two vectors are perpendicular if their dot product is zero. \[ \vec{C} \cdot \vec{D} = 0 \] Step 3: Substitute the expressions for \( \vec{C} \) and \( \vec{D} \) into the dot product equation. \[ (\vec{A} + 2\vec{B}) \cdot (5\vec{A} - 4\vec{B}) = 0 \] Step 4: Expand the dot product using distributive property. \[ \vec{A} \cdot (5\vec{A}) + \vec{A} \cdot (-4\vec{B}) + (2\vec{B}) \cdot (5\vec{A}) + (2\vec{B}) \cdot (-4\vec{B}) = 0 \] \[ 5(\vec{A} \cdot \vec{A}) - 4(\vec{A} \cdot \vec{B}) + 10(\vec{B} \cdot \vec{A}) - 8(\vec{B} \cdot \vec{B}) = 0 \] Step 5: Use the properties of the dot product: \( \vec{X} \cdot \vec{X} = |\vec{X}|^2 \) and \( \vec{X} \cdot \vec{Y} = \vec{Y} \cdot \vec{X} \). \[ 5|\vec{A}|^2 - 4(\vec{A} \cdot \vec{B}) + 10(\vec{A} \cdot \vec{B}) - 8|\vec{B}|^2 = 0 \] \[ 5|\vec{A}|^2 + 6(\vec{A} \cdot \vec{B}) - 8|\vec{B}|^2 = 0 \] Step 6: Substitute the magnitudes of the unit vectors \( |\vec{A}| = 1 \) and \( |\vec{B}| = 1 \). \[ 5(1)^2 + 6(\vec{A} \cdot \vec{B}) - 8(1)^2 = 0 \] \[ 5 + 6(\vec{A} \cdot \vec{B}) - 8 = 0 \] \[ 6(\vec{A} \cdot \vec{B}) - 3 = 0 \] \[ 6(\vec{A} \cdot \vec{B}) = 3 \] \[ \vec{A} \cdot \vec{B} = \frac{3}{6} = \frac{1}{2} \] Step 7: Use the definition of the dot product to find the angle \( \theta \) between \( \vec{A} \) and \( \vec{B} \). \[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \] Substitute the known values: \[ \frac{1}{2} = (1)(1) \cos \theta \] \[ \cos \theta = \frac{1}{2} \] For \( \theta \in [0, \pi] \), the angle is: \[ \theta = \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3} \] Step 8: Compare the result with the given options. The angle between \( \vec{A} \) and \( \vec{B} \) is \( \pi/3 \), which matches option (B).