If `vec(a)=4hat(i)-hat(j)+hat(k)` and `vec(b)=2hat(i)-2hat(j)+hat(k)`, then find a unit vector parallel to the vector `vec(a)+vec(b)`.

Answer

To find a unit vector parallel to the vector \(\vec{a} + \vec{b}\), we will follow these steps:

### Step 1: Find \(\vec{a} + \vec{b}\)

Given:
\[ \vec{a} = 4\hat{i} - \hat{j} + \hat{k} \]
\[ \vec{b} = 2\hat{i} - 2\hat{j} + \hat{k} \]

Now, we add the vectors \(\vec{a}\) and \(\vec{b}\):
\[ \vec{a} + \vec{b} = (4\hat{i} + 2\hat{i}) + (-\hat{j} - 2\hat{j}) + (\hat{k} + \hat{k}) \]
\[ = (4 + 2)\hat{i} + (-1 - 2)\hat{j} + (1 + 1)\hat{k} \]
\[ = 6\hat{i} - 3\hat{j} + 2\hat{k} \]

### Step 2: Calculate the magnitude of \(\vec{a} + \vec{b}\)

The magnitude of a vector \(\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}\) is given by:
\[ |\vec{v}| = \sqrt{x^2 + y^2 + z^2} \]

For \(\vec{a} + \vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}\):
\[ |\vec{a} + \vec{b}| = \sqrt{(6)^2 + (-3)^2 + (2)^2} \]
\[ = \sqrt{36 + 9 + 4} \]
\[ = \sqrt{49} \]
\[ = 7 \]

### Step 3: Find the unit vector parallel to \(\vec{a} + \vec{b}\)

A unit vector in the direction of a vector \(\vec{v}\) is given by:
\[ \hat{u} = \frac{\vec{v}}{|\vec{v}|} \]

Thus, the unit vector parallel to \(\vec{a} + \vec{b}\) is:
\[ \hat{u} = \frac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|} \]
\[ = \frac{6\hat{i} - 3\hat{j} + 2\hat{k}}{7} \]
\[ = \frac{6}{7}\hat{i} - \frac{3}{7}\hat{j} + \frac{2}{7}\hat{k} \]

### Final Answer

The unit vector parallel to \(\vec{a} + \vec{b}\) is:
\[ \hat{u} = \frac{6}{7}\hat{i} - \frac{3}{7}\hat{j} + \frac{2}{7}\hat{k} \]

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