The norm of vector `vecV=3veci-sqrt2vecj` is

Options

A. 4.24
B. 3.61
C. 3.32
D. 2.45

Explanation

To find the norm of the vector \(\vec{V} = 3\vec{i} - \sqrt{2}\vec{j}\), we will follow these steps:

### Step 1: Identify the components of the vector
The vector \(\vec{V}\) can be expressed in terms of its components:
- \(a = 3\) (the coefficient of \(\vec{i}\))
- \(b = -\sqrt{2}\) (the coefficient of \(\vec{j}\))

### Step 2: Use the formula for the norm of a vector
The norm (or magnitude) of a vector \(\vec{V} = a\vec{i} + b\vec{j}\) is given by the formula:
\[ \|\vec{V}\| = \sqrt{a^2 + b^2} \]

### Step 3: Substitute the values into the formula
Substituting \(a = 3\) and \(b = -\sqrt{2}\) into the formula, we get:
\[ \|\vec{V}\| = \sqrt{3^2 + (-\sqrt{2})^2} \]

### Step 4: Calculate \(a^2\) and \(b^2\)
Calculating \(3^2\) and \((- \sqrt{2})^2\):
- \(3^2 = 9\)
- \((- \sqrt{2})^2 = 2\)

### Step 5: Add the squares
Now, we add the results from Step 4:
\[ \|\vec{V}\| = \sqrt{9 + 2} = \sqrt{11} \]

### Step 6: Calculate the square root
Now we need to calculate \(\sqrt{11}\). Using a calculator, we find:
\[ \sqrt{11} \approx 3.3166 \]

### Step 7: Round the result
Rounding to two decimal places, we get:
\[ \|\vec{V}\| \approx 3.32 \]

### Conclusion
Thus, the norm of the vector \(\vec{V} = 3\vec{i} - \sqrt{2}\vec{j}\) is approximately \(3.32\).

### Final Answer
The correct option is \(3.32\).

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