The angle between two vectors `-2hati+3hatj+k` and `hati+2hatj-4hatk` is

Options

A. `45^(@)`
B. `90^(@)`
C. `30^(@)`
D. `60^(@)`

Explanation

To find the angle between the two vectors \(\mathbf{A} = -2\hat{i} + 3\hat{j} + \hat{k}\) and \(\mathbf{B} = \hat{i} + 2\hat{j} - 4\hat{k}\), we will use the dot product formula. The angle \(\theta\) between two vectors can be calculated using the formula:

\[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(\theta) \]

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

The dot product of two vectors \(\mathbf{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) and \(\mathbf{B} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\) is given by:

\[ \mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 + a_3b_3 \]

Substituting the values:

\[ \mathbf{A} \cdot \mathbf{B} = (-2)(1) + (3)(2) + (1)(-4) \]

Calculating each term:

\[ = -2 + 6 - 4 \]

Combining these:

\[ = 0 \]

### Step 2: Use the dot product to find \(\cos(\theta)\)

From the dot product formula, we have:

\[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(\theta) \]

Since we found that \(\mathbf{A} \cdot \mathbf{B} = 0\), we can set up the equation:

\[ 0 = |\mathbf{A}| |\mathbf{B}| \cos(\theta) \]

### Step 3: Analyze the equation

Since the magnitudes \(|\mathbf{A}|\) and \(|\mathbf{B}|\) are both non-zero (as neither vector is the zero vector), we can conclude that:

\[ \cos(\theta) = 0 \]

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

The angle \(\theta\) for which \(\cos(\theta) = 0\) is:

\[ \theta = 90^\circ \]

### Final Answer

Thus, the angle between the two vectors is \(90^\circ\).

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