The angle between the vectors `4hat(i)+3hat(j)-4hat(k)` and `3hat(i)+4hat(j)+6hat(k)` is :

Options

A. `0^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Explanation

To find the angle between the vectors \( \vec{A} = 4\hat{i} + 3\hat{j} - 4\hat{k} \) and \( \vec{B} = 3\hat{i} + 4\hat{j} + 6\hat{k} \), we can use the dot product formula. The angle \( \theta \) between two vectors can be calculated using the formula:

\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \]

### Step 1: Calculate the dot product \( \vec{A} \cdot \vec{B} \)

The dot product \( \vec{A} \cdot \vec{B} \) is calculated as follows:

\[ \vec{A} \cdot \vec{B} = (4)(3) + (3)(4) + (-4)(6) \]

Calculating each term:

- \( 4 \times 3 = 12 \)
- \( 3 \times 4 = 12 \)
- \( -4 \times 6 = -24 \)

Now, summing these values:

\[ \vec{A} \cdot \vec{B} = 12 + 12 - 24 = 0 \]

### Step 2: Calculate the magnitudes \( |\vec{A}| \) and \( |\vec{B}| \)

The magnitude of vector \( \vec{A} \) is given by:

\[ |\vec{A}| = \sqrt{(4^2) + (3^2) + (-4^2)} = \sqrt{16 + 9 + 16} = \sqrt{41} \]

The magnitude of vector \( \vec{B} \) is given by:

\[ |\vec{B}| = \sqrt{(3^2) + (4^2) + (6^2)} = \sqrt{9 + 16 + 36} = \sqrt{61} \]

### Step 3: Substitute values into the cosine formula

Now, substituting the values into the cosine formula:

\[ \cos \theta = \frac{0}{|\vec{A}| |\vec{B}|} = \frac{0}{\sqrt{41} \cdot \sqrt{61}} = 0 \]

### Step 4: Determine the angle \( \theta \)

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

\[ \theta = 90^\circ \]

### Conclusion

The angle between the vectors \( \vec{A} \) and \( \vec{B} \) is \( 90^\circ \).