If `vecA= (2hati + 3hatj + 5hatk)` and `vecB = (hati+ 6hatj+6hatk)`, then projection of `vecA` on `vecB` would be

Options

A. `50/(sqrt73)`
B. `10/(sqrt73)`
C. `40/(sqrt73)`
D. zero

Answer

To find the projection of vector \(\vec{A}\) on vector \(\vec{B}\), we can follow these steps:

### Step 1: Identify the vectors
Given:
\[ \vec{A} = 2\hat{i} + 3\hat{j} + 5\hat{k} \]
\[ \vec{B} = \hat{i} + 6\hat{j} + 6\hat{k} \]

### Step 2: Find the unit vector of \(\vec{B}\)
The unit vector \(\hat{b}\) in the direction of \(\vec{B}\) is given by:
\[ \hat{b} = \frac{\vec{B}}{|\vec{B}|} \]
First, we need to calculate the magnitude of \(\vec{B}\):
\[ |\vec{B}| = \sqrt{(1)^2 + (6)^2 + (6)^2} = \sqrt{1 + 36 + 36} = \sqrt{73} \]
Now, we can find the unit vector:
\[ \hat{b} = \frac{\hat{i} + 6\hat{j} + 6\hat{k}}{\sqrt{73}} \]

### Step 3: Calculate the dot product \(\vec{A} \cdot \hat{b}\)
Now, we find the dot product of \(\vec{A}\) and \(\hat{b}\):
\[ \vec{A} \cdot \hat{b} = (2\hat{i} + 3\hat{j} + 5\hat{k}) \cdot \left(\frac{\hat{i} + 6\hat{j} + 6\hat{k}}{\sqrt{73}}\right) \]
Calculating the dot product:
\[ \vec{A} \cdot \hat{b} = \frac{1}{\sqrt{73}} \left(2 \cdot 1 + 3 \cdot 6 + 5 \cdot 6\right) \]
\[ = \frac{1}{\sqrt{73}} \left(2 + 18 + 30\right) = \frac{50}{\sqrt{73}} \]

### Step 4: Find the projection of \(\vec{A}\) on \(\vec{B}\)
The projection of \(\vec{A}\) on \(\vec{B}\) is given by:
\[ \text{proj}_{\vec{B}} \vec{A} = \left(\vec{A} \cdot \hat{b}\right) \hat{b} \]
Substituting the values:
\[ \text{proj}_{\vec{B}} \vec{A} = \frac{50}{\sqrt{73}} \cdot \frac{\hat{i} + 6\hat{j} + 6\hat{k}}{\sqrt{73}} = \frac{50}{73} (\hat{i} + 6\hat{j} + 6\hat{k}) \]

### Final Answer
Thus, the projection of \(\vec{A}\) on \(\vec{B}\) is:
\[ \text{proj}_{\vec{B}} \vec{A} = \frac{50}{73} \hat{i} + \frac{300}{73} \hat{j} + \frac{300}{73} \hat{k} \]

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