Find the vector component of a vector `2hat(i)+3hat(j)+6hat(k)` along and perpendicular to the non-zero vector `2hat(i)+hat(j)+2hat(k)`.

Answer

To find the vector component of the vector \( \mathbf{a} = 2\hat{i} + 3\hat{j} + 6\hat{k} \) along and perpendicular to the vector \( \mathbf{b} = 2\hat{i} + \hat{j} + 2\hat{k} \), we will follow these steps:

### Step 1: Find the unit vector of \( \mathbf{b} \)

The unit vector \( \hat{b} \) is given by:

\[ \hat{b} = \frac{\mathbf{b}}{|\mathbf{b}|} \]

First, we calculate the magnitude of \( \mathbf{b} \):

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

Now, we can find the unit vector:

\[ \hat{b} = \frac{2\hat{i} + \hat{j} + 2\hat{k}}{3} = \frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k} \]

### Step 2: Find the component of \( \mathbf{a} \) along \( \mathbf{b} \)

The component of \( \mathbf{a} \) along \( \mathbf{b} \) is given by the formula:

\[ \text{Component along } \mathbf{b} = (\mathbf{a} \cdot \hat{b}) \hat{b} \]

First, we compute the dot product \( \mathbf{a} \cdot \hat{b} \):

\[ \mathbf{a} \cdot \hat{b} = (2\hat{i} + 3\hat{j} + 6\hat{k}) \cdot \left(\frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k}\right) \]

Calculating the dot product:

\[ = 2 \cdot \frac{2}{3} + 3 \cdot \frac{1}{3} + 6 \cdot \frac{2}{3} = \frac{4}{3} + 1 + 4 = \frac{4}{3} + \frac{3}{3} + \frac{12}{3} = \frac{19}{3} \]

Now, we can find the component along \( \mathbf{b} \):

\[ \text{Component along } \mathbf{b} = \frac{19}{3} \hat{b} = \frac{19}{3} \left(\frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k}\right) = \frac{38}{9}\hat{i} + \frac{19}{9}\hat{j} + \frac{38}{9}\hat{k} \]

### Step 3: Find the component of \( \mathbf{a} \) perpendicular to \( \mathbf{b} \)

The component of \( \mathbf{a} \) perpendicular to \( \mathbf{b} \) can be found using the formula:

\[ \text{Component perpendicular to } \mathbf{b} = \mathbf{a} - \text{Component along } \mathbf{b} \]

Substituting the values:

\[ \text{Component perpendicular to } \mathbf{b} = (2\hat{i} + 3\hat{j} + 6\hat{k}) - \left(\frac{38}{9}\hat{i} + \frac{19}{9}\hat{j} + \frac{38}{9}\hat{k}\right) \]

Calculating this gives:

\[ = \left(2 - \frac{38}{9}\right)\hat{i} + \left(3 - \frac{19}{9}\right)\hat{j} + \left(6 - \frac{38}{9}\right)\hat{k} \]

Converting the constants to fractions:

\[ = \left(\frac{18}{9} - \frac{38}{9}\right)\hat{i} + \left(\frac{27}{9} - \frac{19}{9}\right)\hat{j} + \left(\frac{54}{9} - \frac{38}{9}\right)\hat{k} \]

This simplifies to:

\[ = \left(-\frac{20}{9}\right)\hat{i} + \left(\frac{8}{9}\right)\hat{j} + \left(\frac{16}{9}\right)\hat{k} \]

### Final Result

Thus, the components are:

- **Component of \( \mathbf{a} \) along \( \mathbf{b} \)**: \( \frac{38}{9}\hat{i} + \frac{19}{9}\hat{j} + \frac{38}{9}\hat{k} \)
- **Component of \( \mathbf{a} \) perpendicular to \( \mathbf{b} \)**: \( -\frac{20}{9}\hat{i} + \frac{8}{9}\hat{j} + \frac{16}{9}\hat{k} \)