If `veca, vecb and vecc` are three vectors, such that `|veca|=2, |vecb|=3, |vecc|=4, veca. vecc=0, veca. vecb=0` and the angle between `vecb and vecc` is `(pi)/(3)`, then the value of `|veca xx (2vecb - 3 vecc)|` is equal to

Options

A. `12sqrt3`
B. `6sqrt3`
C. `3sqrt3`
D. 5

Answer

To solve the problem, we need to find the value of \(|\vec{a} \times (2\vec{b} - 3\vec{c})|\). We will follow these steps:

### Step 1: Calculate the Magnitude of the Vector \(2\vec{b} - 3\vec{c}\)

We know the magnitudes of the vectors:
- \(|\vec{a}| = 2\)
- \(|\vec{b}| = 3\)
- \(|\vec{c}| = 4\)

We also know that \(\vec{a} \cdot \vec{b} = 0\) and \(\vec{a} \cdot \vec{c} = 0\), which means \(\vec{a}\) is orthogonal to both \(\vec{b}\) and \(\vec{c}\).

The angle between \(\vec{b}\) and \(\vec{c}\) is \(\frac{\pi}{3}\) (or 60 degrees). We can calculate the dot product \(\vec{b} \cdot \vec{c}\) using the formula:
\[ \vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos\theta = 3 \cdot 4 \cdot \cos\left(\frac{\pi}{3}\right) = 12 \cdot \frac{1}{2} = 6 \]

Now, we can find the magnitude of the vector \(2\vec{b} - 3\vec{c}\):
\[ |2\vec{b} - 3\vec{c}|^2 = |2\vec{b}|^2 + |-3\vec{c}|^2 - 2(2\vec{b}) \cdot (-3\vec{c}) \]
Calculating each term:
- \(|2\vec{b}|^2 = (2 \cdot 3)^2 = 36\)
- \(|-3\vec{c}|^2 = (3 \cdot 4)^2 = 144\)
- \(2(2\vec{b}) \cdot (-3\vec{c}) = -12 \cdot 6 = -72\)

Putting it all together:
\[ |2\vec{b} - 3\vec{c}|^2 = 36 + 144 + 72 = 252 \]
Thus,
\[ |2\vec{b} - 3\vec{c}| = \sqrt{252} = 6\sqrt{7} \]

### Step 2: Calculate the Angle Between \(\vec{a}\) and \(2\vec{b} - 3\vec{c}\)

Since \(\vec{a}\) is orthogonal to both \(\vec{b}\) and \(\vec{c}\), the angle between \(\vec{a}\) and \(2\vec{b} - 3\vec{c}\) can be found using the dot product:
\[ \vec{a} \cdot (2\vec{b} - 3\vec{c}) = 2(\vec{a} \cdot \vec{b}) - 3(\vec{a} \cdot \vec{c}) = 0 - 0 = 0 \]
This means that the angle \(\theta\) between \(\vec{a}\) and \(2\vec{b} - 3\vec{c}\) is \(\frac{\pi}{2}\) (90 degrees).

### Step 3: Calculate the Magnitude of the Cross Product

The magnitude of the cross product is given by:
\[ |\vec{a} \times (2\vec{b} - 3\vec{c})| = |\vec{a}| |2\vec{b} - 3\vec{c}| \sin\theta \]
Substituting the values we have:
- \(|\vec{a}| = 2\)
- \(|2\vec{b} - 3\vec{c}| = 6\sqrt{7}\)
- \(\sin\left(\frac{\pi}{2}\right) = 1\)

Thus,
\[ |\vec{a} \times (2\vec{b} - 3\vec{c})| = 2 \cdot 6\sqrt{7} \cdot 1 = 12\sqrt{7} \]

### Final Answer
The value of \(|\vec{a} \times (2\vec{b} - 3\vec{c})|\) is \(12\sqrt{7}\).

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