If `veca and vecb ` are any two vectors of `magnitude` `2 and 3` respectively such that `|2(vecaxxvecb)|+|3(veca.vecb)|=k` then the maximum value of k is

Options

A. `sqrt13`
B. `2sqrt13`
C. `6sqrt13`
D. `10sqrt13`

Explanation

To find the maximum value of \( k \) given the equation \( |2(\vec{a} \times \vec{b})| + |3(\vec{a} \cdot \vec{b})| = k \), where the magnitudes of vectors \( \vec{a} \) and \( \vec{b} \) are \( 2 \) and \( 3 \) respectively, we can follow these steps:

### Step 1: Use the identities for cross product and dot product
We know the following identities:
- The magnitude of the cross product:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \]
- The dot product:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]

### Step 2: Substitute the magnitudes of the vectors
Given \( |\vec{a}| = 2 \) and \( |\vec{b}| = 3 \), we can substitute these values into the identities:
- For the cross product:
\[ |\vec{a} \times \vec{b}| = 2 \cdot 3 \cdot \sin \theta = 6 \sin \theta \]
- For the dot product:
\[ \vec{a} \cdot \vec{b} = 2 \cdot 3 \cdot \cos \theta = 6 \cos \theta \]

### Step 3: Substitute into the equation for \( k \)
Now substituting these results into the equation for \( k \):
\[ k = |2(\vec{a} \times \vec{b})| + |3(\vec{a} \cdot \vec{b})| \]
This becomes:
\[ k = |2 \cdot 6 \sin \theta| + |3 \cdot 6 \cos \theta| \]
\[ k = 12 |\sin \theta| + 18 |\cos \theta| \]

### Step 4: Maximize \( k \)
To maximize \( k \), we can express it in terms of a single variable. We can factor out the common term:
\[ k = 6(2 |\sin \theta| + 3 |\cos \theta|) \]
Let \( x = |\sin \theta| \) and \( y = |\cos \theta| \). We know that \( x^2 + y^2 = 1 \).

### Step 5: Use the Cauchy-Schwarz inequality
To find the maximum of \( 2x + 3y \) subject to \( x^2 + y^2 = 1 \), we can apply the Cauchy-Schwarz inequality:
\[ (2^2 + 3^2)(x^2 + y^2) \geq (2x + 3y)^2 \]
This simplifies to:
\[ (4 + 9)(1) \geq (2x + 3y)^2 \]
\[ 13 \geq (2x + 3y)^2 \]
Taking the square root gives:
\[ \sqrt{13} \geq 2x + 3y \]

### Step 6: Find the maximum value of \( k \)
Thus, the maximum value of \( 2x + 3y \) is \( \sqrt{13} \). Therefore, the maximum value of \( k \) is:
\[ k_{\text{max}} = 6 \sqrt{13} \]

### Final Answer
The maximum value of \( k \) is \( 6\sqrt{13} \).

---