A parallelogram is constructed on `3veca+vecb and veca-4vecb, where |veca|=6 and |vecb|=8` and `veca and vecb` are anti parallel then the length of the longer diagonal is (A) 40 (B) 64 (C) 32 (D) 48

Options

A. 40
B. 64
C. 32
D. 48

Explanation

To solve the problem, we need to find the length of the longer diagonal of the parallelogram constructed on the vectors \( \vec{u} = 3\vec{a} + \vec{b} \) and \( \vec{v} = \vec{a} - 4\vec{b} \). We are given that \( |\vec{a}| = 6 \) and \( |\vec{b}| = 8 \), and that \( \vec{a} \) and \( \vec{b} \) are anti-parallel.

### Step-by-Step Solution:

1. **Identify the Vectors**:
\[ \vec{u} = 3\vec{a} + \vec{b} \]
\[ \vec{v} = \vec{a} - 4\vec{b} \]

2. **Calculate the Diagonal Vectors**:
The diagonals of the parallelogram can be found using the formula:
\[ \text{Diagonal 1} = \vec{u} + \vec{v} \]
\[ \text{Diagonal 2} = \vec{u} - \vec{v} \]

Let's calculate \( \vec{u} + \vec{v} \):
\[ \vec{u} + \vec{v} = (3\vec{a} + \vec{b}) + (\vec{a} - 4\vec{b}) = 4\vec{a} - 3\vec{b} \]

3. **Find the Magnitude of the Diagonal**:
We need to calculate the magnitude of \( 4\vec{a} - 3\vec{b} \):
\[ |4\vec{a} - 3\vec{b}|^2 = |4\vec{a}|^2 + |-3\vec{b}|^2 - 2(4\vec{a}) \cdot (-3\vec{b}) \]

Using the magnitudes:
\[ |4\vec{a}|^2 = 16|\vec{a}|^2 = 16 \times 36 = 576 \]
\[ |-3\vec{b}|^2 = 9|\vec{b}|^2 = 9 \times 64 = 576 \]

Now, calculate the dot product \( (4\vec{a}) \cdot (-3\vec{b}) \):
Since \( \vec{a} \) and \( \vec{b} \) are anti-parallel, \( \cos \theta = -1 \):
\[ (4\vec{a}) \cdot (-3\vec{b}) = -12|\vec{a}||\vec{b}| = -12 \times 6 \times 8 = -576 \]

Now substituting back:
\[ |4\vec{a} - 3\vec{b}|^2 = 576 + 576 + 2 \times 576 = 576 + 576 + 1152 = 2304 \]

4. **Calculate the Magnitude**:
\[ |4\vec{a} - 3\vec{b}| = \sqrt{2304} = 48 \]

Thus, the length of the longer diagonal is \( \boxed{48} \).