The projection of a vector `vec(r )=3hat(i)+hat(j)+2hat(k)` on the `x-y` plane has magnitude

Options

A. 3
B. 4
C. `sqrt(14)`
D. `sqrt(10)`

Explanation

To find the magnitude of the projection of the vector \(\vec{r} = 3\hat{i} + \hat{j} + 2\hat{k}\) on the \(x-y\) plane, we can follow these steps:

### Step 1: Identify the components of the vector
The vector \(\vec{r}\) can be broken down into its components:
- \(x\) component: \(3\) (from \(3\hat{i}\))
- \(y\) component: \(1\) (from \(\hat{j}\))
- \(z\) component: \(2\) (from \(2\hat{k}\))

### Step 2: Determine the projection on the \(x-y\) plane
The projection of the vector on the \(x-y\) plane only considers the \(x\) and \(y\) components. Therefore, the projection of \(\vec{r}\) on the \(x-y\) plane is:
\[ \vec{r}_{xy} = 3\hat{i} + \hat{j} \]

### Step 3: Calculate the magnitude of the projection
The magnitude of a vector \(\vec{v} = a\hat{i} + b\hat{j}\) is given by:
\[ |\vec{v}| = \sqrt{a^2 + b^2} \]
For our projection \(\vec{r}_{xy} = 3\hat{i} + 1\hat{j}\):
- \(a = 3\)
- \(b = 1\)

Now, we can calculate the magnitude:
\[ |\vec{r}_{xy}| = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \]

### Step 4: Conclusion
The magnitude of the projection of the vector \(\vec{r}\) on the \(x-y\) plane is \(\sqrt{10}\).

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