The sine of the angle between `3hat(i)+hat(j)+2hat(k)and 2hat(i)-2hat(j)+4hat(k)` is

Options

A. 1
B. 0.91
C. 0.76
D. 0.67

Answer

To find the sine of the angle between the vectors \( \mathbf{A} = 3\hat{i} + \hat{j} + 2\hat{k} \) and \( \mathbf{B} = 2\hat{i} - 2\hat{j} + 4\hat{k} \), we can follow these steps:

### Step 1: Calculate the dot product of the two vectors
The dot product \( \mathbf{A} \cdot \mathbf{B} \) is given by:
\[ \mathbf{A} \cdot \mathbf{B} = (3)(2) + (1)(-2) + (2)(4) \]
Calculating this gives:
\[ \mathbf{A} \cdot \mathbf{B} = 6 - 2 + 8 = 12 \]

### Step 2: Calculate the magnitudes of the vectors
The magnitude of vector \( \mathbf{A} \) is:
\[ |\mathbf{A}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \]
The magnitude of vector \( \mathbf{B} \) is:
\[ |\mathbf{B}| = \sqrt{2^2 + (-2)^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} \]

### Step 3: Use the dot product to find cos(θ)
From the dot product formula:
\[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta \]
We can rearrange this to find \( \cos \theta \):
\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} = \frac{12}{\sqrt{14} \cdot \sqrt{24}} \]

### Step 4: Simplify the expression for cos(θ)
Calculating \( \sqrt{14} \cdot \sqrt{24} \):
\[ \sqrt{14 \cdot 24} = \sqrt{336} \]
Thus,
\[ \cos \theta = \frac{12}{\sqrt{336}} = \frac{12}{\sqrt{16 \cdot 21}} = \frac{12}{4\sqrt{21}} = \frac{3}{\sqrt{21}} \]

### Step 5: Find sin(θ) using the identity sin²(θ) + cos²(θ) = 1
We know:
\[ \sin^2 \theta = 1 - \cos^2 \theta \]
Calculating \( \cos^2 \theta \):
\[ \cos^2 \theta = \left(\frac{3}{\sqrt{21}}\right)^2 = \frac{9}{21} = \frac{3}{7} \]
Now substituting into the identity:
\[ \sin^2 \theta = 1 - \frac{3}{7} = \frac{4}{7} \]
Thus,
\[ \sin \theta = \sqrt{\frac{4}{7}} = \frac{2}{\sqrt{7}} \]

### Step 6: Final answer
The sine of the angle between the two vectors is:
\[ \sin \theta = \frac{2}{\sqrt{7}} \approx 0.7559 \]