Find angle between the vectors `2i+j+k" and "i-j+k`.

Answer

To find the angle between the vectors \( \mathbf{a} = 2\mathbf{i} + \mathbf{j} + \mathbf{k} \) and \( \mathbf{b} = \mathbf{i} - \mathbf{j} + \mathbf{k} \), 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 \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \) and \( \mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k} \) is given by:

\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]

For our vectors:

\[ \mathbf{a} = 2\mathbf{i} + 1\mathbf{j} + 1\mathbf{k} \quad \text{and} \quad \mathbf{b} = 1\mathbf{i} - 1\mathbf{j} + 1\mathbf{k} \]

Calculating the dot product:

\[ \mathbf{a} \cdot \mathbf{b} = (2)(1) + (1)(-1) + (1)(1) = 2 - 1 + 1 = 2 \]

### Step 2: Calculate the magnitudes \( |\mathbf{a}| \) and \( |\mathbf{b}| \)

The magnitude of a vector \( \mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k} \) is given by:

\[ |\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]

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

\[ |\mathbf{a}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]

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

\[ |\mathbf{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

### Step 3: Substitute into the cosine formula

Now we can substitute the dot product and the magnitudes into the cosine formula:

\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} = \frac{2}{\sqrt{6} \cdot \sqrt{3}} = \frac{2}{\sqrt{18}} = \frac{2}{3\sqrt{2}} \]

### Step 4: Simplify the expression

To simplify \( \frac{2}{3\sqrt{2}} \):

\[ \cos \theta = \frac{2\sqrt{2}}{6} = \frac{\sqrt{2}}{3} \]

### Step 5: Find the angle \( \theta \)

To find the angle \( \theta \), we take the inverse cosine:

\[ \theta = \cos^{-1}\left(\frac{\sqrt{2}}{3}\right) \]

### Final Answer

Thus, the angle between the vectors \( 2\mathbf{i} + \mathbf{j} + \mathbf{k} \) and \( \mathbf{i} - \mathbf{j} + \mathbf{k} \) is:

\[ \theta = \cos^{-1}\left(\frac{\sqrt{2}}{3}\right) \]