The angle between the vectors i-j+k and -i+j+2k is

Options

A. `45^(@)`
B. `60^(@)`
C. `90^(@)`
D. `135^(@)`

Answer

To find the angle between the vectors \( \mathbf{a} = \mathbf{i} - \mathbf{j} + \mathbf{k} \) and \( \mathbf{b} = -\mathbf{i} + \mathbf{j} + 2\mathbf{k} \), we can use the formula involving the dot product of the vectors.

### Step 1: Calculate the dot product of the vectors
The dot product \( \mathbf{a} \cdot \mathbf{b} \) is given by:
\[ \mathbf{a} \cdot \mathbf{b} = (1)(-1) + (-1)(1) + (1)(2) \]
Calculating this gives:
\[ \mathbf{a} \cdot \mathbf{b} = -1 - 1 + 2 = 0 \]

### Step 2: Calculate the magnitudes of the vectors
The magnitude of vector \( \mathbf{a} \) is:
\[ |\mathbf{a}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

The magnitude of vector \( \mathbf{b} \) is:
\[ |\mathbf{b}| = \sqrt{(-1)^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]

### Step 3: Use the dot product to find the cosine of the angle
The formula relating the dot product and the angle \( \theta \) between two vectors is:
\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \]
Substituting the known values:
\[ 0 = \sqrt{3} \cdot \sqrt{6} \cdot \cos \theta \]

### Step 4: Solve for \( \cos \theta \)
Since the left-hand side is zero:
\[ 0 = \sqrt{3} \cdot \sqrt{6} \cdot \cos \theta \]
This implies that:
\[ \cos \theta = 0 \]

### Step 5: Determine the angle \( \theta \)
The cosine of an angle is zero at \( \theta = \frac{\pi}{2} \) or \( 90^\circ \). Therefore, the angle between the vectors is:
\[ \theta = 90^\circ \]

### Final Answer
The angle between the vectors \( \mathbf{i} - \mathbf{j} + \mathbf{k} \) and \( -\mathbf{i} + \mathbf{j} + 2\mathbf{k} \) is \( 90^\circ \).

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