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

Options

A. `0^(@)`
B. `90^(@)`
C. `180^(@)`
D. None of these

Answer

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 can use the formula for the cosine of the angle \(\theta\) between two vectors:

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

### 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:

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

For our vectors:

- \(a_1 = -2\), \(a_2 = 3\), \(a_3 = 1\)
- \(b_1 = 1\), \(b_2 = 2\), \(b_3 = -4\)

Calculating the dot product:

\[ \mathbf{A} \cdot \mathbf{B} = (-2)(1) + (3)(2) + (1)(-4) = -2 + 6 - 4 = 0 \]

### Step 2: Calculate the magnitudes of \(\mathbf{A}\) and \(\mathbf{B}\)

The magnitude of a vector \(\mathbf{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) is given by:

\[ |\mathbf{A}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]

Calculating the magnitude of \(\mathbf{A}\):

\[ |\mathbf{A}| = \sqrt{(-2)^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14} \]

Calculating the magnitude of \(\mathbf{B}\):

\[ |\mathbf{B}| = \sqrt{(1)^2 + (2)^2 + (-4)^2} = \sqrt{1 + 4 + 16} = \sqrt{21} \]

### Step 3: Substitute into the cosine formula

Now we can substitute the values into the cosine formula:

\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} = \frac{0}{\sqrt{14} \cdot \sqrt{21}} = 0 \]

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

Since \(\cos \theta = 0\), we find that:

\[ \theta = 90^\circ \]

### Final Answer

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

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