Find the scalar projection of :
`vec(a)=3hat(i)-2hat(j)+hat(k)` on `vec(b)=hat(i)-2hat(j)-3hat(k)`

Answer

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

### Step 1: Identify the vectors
Given:
\[ \vec{a} = 3\hat{i} - 2\hat{j} + \hat{k} \]
\[ \vec{b} = \hat{i} - 2\hat{j} - 3\hat{k} \]

### Step 2: Find the unit vector of \(\vec{b}\)
To find the unit vector of \(\vec{b}\), we first need to calculate the magnitude of \(\vec{b}\).

The magnitude of \(\vec{b}\) is given by:
\[ |\vec{b}| = \sqrt{(1)^2 + (-2)^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]

Now, the unit vector \(\hat{b}\) is:
\[ \hat{b} = \frac{\vec{b}}{|\vec{b}|} = \frac{\hat{i} - 2\hat{j} - 3\hat{k}}{\sqrt{14}} \]

### Step 3: Calculate the dot product \(\vec{a} \cdot \hat{b}\)
Now we can calculate the dot product \(\vec{a} \cdot \hat{b}\):
\[ \vec{a} \cdot \vec{b} = (3\hat{i} - 2\hat{j} + \hat{k}) \cdot (\hat{i} - 2\hat{j} - 3\hat{k}) \]
Calculating the dot product:
\[ = 3(1) + (-2)(-2) + (1)(-3) \]
\[ = 3 + 4 - 3 = 4 \]

### Step 4: Find the scalar projection of \(\vec{a}\) on \(\vec{b}\)
The scalar projection of \(\vec{a}\) on \(\vec{b}\) is given by:
\[ \text{Scalar Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \]
Substituting the values we have:
\[ \text{Scalar Projection} = \frac{4}{\sqrt{14}} \]

### Step 5: Final result
Thus, the scalar projection of \(\vec{a}\) on \(\vec{b}\) is:
\[ \frac{4}{\sqrt{14}} \]