Find the angle between the vectors :
`vec(a)=3vec(i)-2vec(j)+vec(k)` and `vec(b)=vec(i)-2vec(j)-3vec(k)`

Answer

To find the angle between the vectors \(\vec{a} = 3\vec{i} - 2\vec{j} + \vec{k}\) and \(\vec{b} = \vec{i} - 2\vec{j} - 3\vec{k}\), we can use the formula for the cosine of the angle \(\theta\) between two vectors:

\[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \]

### Step 1: Calculate the dot product \(\vec{a} \cdot \vec{b}\)

The dot product of two vectors \(\vec{a} = a_1 \vec{i} + a_2 \vec{j} + a_3 \vec{k}\) and \(\vec{b} = b_1 \vec{i} + b_2 \vec{j} + b_3 \vec{k}\) is given by:

\[ \vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \]

For our vectors:
- \(\vec{a} = 3\vec{i} - 2\vec{j} + 1\vec{k}\) (where \(a_1 = 3\), \(a_2 = -2\), \(a_3 = 1\))
- \(\vec{b} = 1\vec{i} - 2\vec{j} - 3\vec{k}\) (where \(b_1 = 1\), \(b_2 = -2\), \(b_3 = -3\))

Calculating the dot product:

\[ \vec{a} \cdot \vec{b} = (3)(1) + (-2)(-2) + (1)(-3) = 3 + 4 - 3 = 4 \]

### Step 2: Calculate the magnitudes of \(\vec{a}\) and \(\vec{b}\)

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

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

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

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

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

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

### Step 3: Substitute into the cosine formula

Now we can substitute the values into the cosine formula:

\[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{4}{\sqrt{14} \cdot \sqrt{14}} = \frac{4}{14} = \frac{2}{7} \]

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

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

\[ \theta = \cos^{-1}\left(\frac{2}{7}\right) \]

### Final Answer

Thus, the angle between the vectors \(\vec{a}\) and \(\vec{b}\) is:

\[ \theta = \cos^{-1}\left(\frac{2}{7}\right) \]

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