Let `veca and vecb` are vectors such that `|veca|=2, |vecb|=3 and veca. vecb=4`. If `vecc=(3veca xx vecb)-4vecb`, then `|vecc|` is equal to

Answer

To solve the problem, we need to find the magnitude of the vector \(\vec{c}\), defined as:

\[ \vec{c} = 3(\vec{a} \times \vec{b}) - 4\vec{b} \]

We are given the following information:

- \(|\vec{a}| = 2\)
- \(|\vec{b}| = 3\)
- \(\vec{a} \cdot \vec{b} = 4\)

### Step 1: Calculate \(|\vec{a} \times \vec{b}|\)

Using the property of dot and cross products, we know:

\[ |\vec{a}|^2 + |\vec{b}|^2 = |\vec{a} \cdot \vec{b}|^2 + |\vec{a} \times \vec{b}|^2 \]

Substituting the known values:

\[ 2^2 + 3^2 = 4^2 + |\vec{a} \times \vec{b}|^2 \]

Calculating the squares:

\[ 4 + 9 = 16 + |\vec{a} \times \vec{b}|^2 \]

This simplifies to:

\[ 13 = 16 + |\vec{a} \times \vec{b}|^2 \]

Rearranging gives:

\[ |\vec{a} \times \vec{b}|^2 = 13 - 16 = -3 \]

This is incorrect, so let's recalculate:

\[ |\vec{a}|^2 + |\vec{b}|^2 = 4 + 9 = 13 \]

And we know:

\[ |\vec{a} \cdot \vec{b}|^2 = 4^2 = 16 \]

Thus:

\[ |\vec{a} \times \vec{b}|^2 = 13 - 16 = -3 \]

This indicates an error in the calculations. Let's correct it.

### Step 2: Correct Calculation of \(|\vec{a} \times \vec{b}|\)

Using the correct relation:

\[ |\vec{a}|^2 + |\vec{b}|^2 = |\vec{a} \cdot \vec{b}|^2 + |\vec{a} \times \vec{b}|^2 \]

Substituting the values:

\[ 4 + 9 = 16 + |\vec{a} \times \vec{b}|^2 \]

This simplifies to:

\[ 13 = 16 + |\vec{a} \times \vec{b}|^2 \]

So, we have:

\[ |\vec{a} \times \vec{b}|^2 = 13 - 16 = -3 \]

This indicates an error in the calculations.

### Step 3: Calculate \(|\vec{c}|\)

Now, we will calculate \(|\vec{c}|\):

\[ |\vec{c}|^2 = |3(\vec{a} \times \vec{b}) - 4\vec{b}|^2 \]

Using the formula for the magnitude of a vector subtraction:

\[ |\vec{c}|^2 = |3(\vec{a} \times \vec{b})|^2 + |-4\vec{b}|^2 - 2 \cdot |3(\vec{a} \times \vec{b})| \cdot |-4\vec{b}| \cdot \cos(\theta) \]

Where \(\theta\) is the angle between \(\vec{a} \times \vec{b}\) and \(\vec{b}\).

### Step 4: Substitute Values

Now substituting values:

1. \(|3(\vec{a} \times \vec{b})|^2 = 9 |\vec{a} \times \vec{b}|^2\)
2. \(|-4\vec{b}|^2 = 16 |\vec{b}|^2 = 16 \cdot 9 = 144\)

Now we need \(|\vec{a} \times \vec{b}|\):

Using the correct calculations, we find:

\[ |\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 \]

Substituting:

\[ |\vec{a} \times \vec{b}|^2 = 2^2 \cdot 3^2 - 4^2 = 4 \cdot 9 - 16 = 36 - 16 = 20 \]

Thus:

\[ |\vec{a} \times \vec{b}| = \sqrt{20} = 2\sqrt{5} \]

### Step 5: Final Calculation

Now substituting back into \(|\vec{c}|^2\):

\[ |\vec{c}|^2 = 9 \cdot 20 + 144 - 2 \cdot 3 \cdot 4 \cdot \cos(\theta) \]

Assuming \(\theta = 90^\circ\) (since \(\vec{a} \times \vec{b}\) is perpendicular to both):

\[ |\vec{c}|^2 = 180 + 144 = 324 \]

Thus:

\[ |\vec{c}| = \sqrt{324} = 18 \]

### Conclusion

Therefore, the magnitude of \(\vec{c}\) is:

\[ |\vec{c}| = 18 \]