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 find the magnitude of the vector \(\vec{c}\) defined as \(\vec{c} = 3(\vec{a} \times \vec{b}) - 4\vec{b}\), we will follow these steps:

### Step 1: Calculate the magnitude of \(\vec{a} \times \vec{b}\)

The magnitude of the cross product of two vectors can be calculated using the formula:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \]
where \(\theta\) is the angle between the vectors \(\vec{a}\) and \(\vec{b}\).

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

We can find \(\cos \theta\) using the dot product formula:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]
Substituting the known values:
\[ 4 = 2 \cdot 3 \cdot \cos \theta \implies 4 = 6 \cos \theta \implies \cos \theta = \frac{2}{3} \]

Now, we can find \(\sin \theta\) using the identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
\[ \sin^2 \theta = 1 - \left(\frac{2}{3}\right)^2 = 1 - \frac{4}{9} = \frac{5}{9} \implies \sin \theta = \frac{\sqrt{5}}{3} \]

Now substituting back to find \(|\vec{a} \times \vec{b}|\):
\[ |\vec{a} \times \vec{b}| = 2 \cdot 3 \cdot \frac{\sqrt{5}}{3} = 2\sqrt{5} \]

### Step 2: Calculate \(|\vec{c}|\)

Now we can substitute \(|\vec{a} \times \vec{b}|\) into the expression for \(\vec{c}\):
\[ \vec{c} = 3(\vec{a} \times \vec{b}) - 4\vec{b} \]

To find \(|\vec{c}|\), we use the formula for the magnitude of the difference of two vectors:
\[ |\vec{c}|^2 = |3(\vec{a} \times \vec{b}) - 4\vec{b}|^2 \]
Using the property of magnitudes:
\[ |\vec{c}|^2 = |3(\vec{a} \times \vec{b})|^2 + |-4\vec{b}|^2 - 2 \cdot |3(\vec{a} \times \vec{b})| \cdot |-4\vec{b}| \cdot \cos \phi \]
where \(\phi\) is the angle between \(3(\vec{a} \times \vec{b})\) and \(-4\vec{b}\). Since \(\vec{a} \times \vec{b}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\), \(\phi = 90^\circ\) and \(\cos \phi = 0\).

Calculating the magnitudes:
\[ |3(\vec{a} \times \vec{b})| = 3 \cdot 2\sqrt{5} = 6\sqrt{5} \]
\[ |-4\vec{b}| = 4 \cdot 3 = 12 \]

Now substituting back:
\[ |\vec{c}|^2 = (6\sqrt{5})^2 + (12)^2 \]
\[ = 180 + 144 = 324 \]

Finally, taking the square root gives us:
\[ |\vec{c}| = \sqrt{324} = 18 \]

### Final Answer
Thus, the magnitude of \(\vec{c}\) is \(\boxed{18}\).