Find the projection of the vector `vec(P) = 2hat(i) - 3hat(j) + 6 hat(k) ` on the vector `vec(Q) = hat(i) + 2hat(j) + 2hat(k)`

Explanation

To find the projection of the vector \(\vec{P} = 2\hat{i} - 3\hat{j} + 6\hat{k}\) on the vector \(\vec{Q} = \hat{i} + 2\hat{j} + 2\hat{k}\), we can use the formula for the projection of one vector onto another:

\[ \text{Projection of } \vec{P} \text{ on } \vec{Q} = \frac{\vec{P} \cdot \vec{Q}}{|\vec{Q}|^2} \vec{Q} \]

### Step 1: Calculate the dot product \(\vec{P} \cdot \vec{Q}\)

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

\[ \vec{A} \cdot \vec{B} = a_1b_1 + a_2b_2 + a_3b_3 \]

For \(\vec{P} = 2\hat{i} - 3\hat{j} + 6\hat{k}\) and \(\vec{Q} = \hat{i} + 2\hat{j} + 2\hat{k}\):

\[ \vec{P} \cdot \vec{Q} = (2)(1) + (-3)(2) + (6)(2) \]
\[ = 2 - 6 + 12 = 8 \]

### Step 2: Calculate the magnitude of \(\vec{Q}\)

The magnitude of a vector \(\vec{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) is given by:

\[ |\vec{A}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]

For \(\vec{Q} = \hat{i} + 2\hat{j} + 2\hat{k}\):

\[ |\vec{Q}| = \sqrt{1^2 + 2^2 + 2^2} \]
\[ = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

### Step 3: Calculate the projection of \(\vec{P}\) on \(\vec{Q}\)

Now we can find the projection using the formula:

\[ \text{Projection of } \vec{P} \text{ on } \vec{Q} = \frac{\vec{P} \cdot \vec{Q}}{|\vec{Q}|^2} \vec{Q} \]

First, we need \( |\vec{Q}|^2 \):

\[ |\vec{Q}|^2 = 3^2 = 9 \]

Now substituting the values:

\[ \text{Projection of } \vec{P} \text{ on } \vec{Q} = \frac{8}{9} \vec{Q} \]
\[ = \frac{8}{9} (\hat{i} + 2\hat{j} + 2\hat{k}) \]

### Final Result

Thus, the projection of \(\vec{P}\) on \(\vec{Q}\) is:

\[ \frac{8}{9} \hat{i} + \frac{16}{9} \hat{j} + \frac{16}{9} \hat{k} \]