Angle between the vectors `sqrt3(veca xx vec b) " and " vec b - (veca .vecb)veca` is

Options

A. `pi/2`
B. 0
C. `pi/4`
D. `pi/3`

Answer

To find the angle between the vectors \( \sqrt{3} (\vec{a} \times \vec{b}) \) and \( \vec{b} - (\vec{a} \cdot \vec{b}) \vec{a} \), we can use the formula for the cosine of the angle between two vectors, which is given by:

\[ \cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} \]

where \( \vec{u} = \sqrt{3} (\vec{a} \times \vec{b}) \) and \( \vec{v} = \vec{b} - (\vec{a} \cdot \vec{b}) \vec{a} \).

### Step 1: Calculate \( \vec{u} \cdot \vec{v} \)

First, we need to compute the dot product \( \vec{u} \cdot \vec{v} \):

\[ \vec{u} \cdot \vec{v} = \sqrt{3} (\vec{a} \times \vec{b}) \cdot \left( \vec{b} - (\vec{a} \cdot \vec{b}) \vec{a} \right) \]

Using the distributive property of the dot product, we can expand this:

\[ \vec{u} \cdot \vec{v} = \sqrt{3} \left[ (\vec{a} \times \vec{b}) \cdot \vec{b} - (\vec{a} \cdot \vec{b}) (\vec{a} \times \vec{b}) \cdot \vec{a} \right] \]

### Step 2: Simplify the terms

1. The term \( (\vec{a} \times \vec{b}) \cdot \vec{b} \) is zero because the cross product \( \vec{a} \times \vec{b} \) is perpendicular to both \( \vec{a} \) and \( \vec{b} \).

Thus, \( (\vec{a} \times \vec{b}) \cdot \vec{b} = 0 \).

2. The term \( (\vec{a} \cdot \vec{b}) (\vec{a} \times \vec{b}) \cdot \vec{a} \) is also zero because \( \vec{a} \times \vec{b} \) is perpendicular to \( \vec{a} \).

Thus, \( (\vec{a} \times \vec{b}) \cdot \vec{a} = 0 \).

Putting these together, we have:

\[ \vec{u} \cdot \vec{v} = \sqrt{3} \left[ 0 - 0 \right] = 0 \]

### Step 3: Calculate the magnitudes of \( \vec{u} \) and \( \vec{v} \)

1. The magnitude of \( \vec{u} \):

\[ |\vec{u}| = |\sqrt{3} (\vec{a} \times \vec{b})| = \sqrt{3} |\vec{a} \times \vec{b}| \]

Using the formula for the magnitude of the cross product:

\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \]

where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \).

Thus,

\[ |\vec{u}| = \sqrt{3} |\vec{a}| |\vec{b}| \sin \theta \]

2. The magnitude of \( \vec{v} \):

\[ |\vec{v}| = |\vec{b} - (\vec{a} \cdot \vec{b}) \vec{a}| \]

Using the projection formula, we can find:

\[ |\vec{v}| = |\vec{b}| \sqrt{1 - \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}\right)^2} \]

### Step 4: Find \( \cos \theta \)

Now we can substitute back into the cosine formula:

\[ \cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} = \frac{0}{|\vec{u}| |\vec{v}|} = 0 \]

### Conclusion

Since \( \cos \theta = 0 \), this implies that:

\[ \theta = 90^\circ \]

Thus, the angle between the vectors \( \sqrt{3} (\vec{a} \times \vec{b}) \) and \( \vec{b} - (\vec{a} \cdot \vec{b}) \vec{a} \) is \( 90^\circ \).