Question

If the area of the parallelogram with $\vec{a}$ and $\vec{b}$ as two adjacent sides is $15$ s units then the area of the parallelogram having $3 \vec{a}+2 \vec{b}$ and $\vec{a}+3 \vec{b}$ as two adjacent sides in s units is

• A. 45
• B. 75
• C. 105
• D. 120

Answer 1

The Correct Option is C

To find the area of the parallelogram having \(3\vec{a} + 2\vec{b}\) and \(\vec{a} + 3\vec{b}\) as two adjacent sides, we need to use the properties of vector cross products to find the area formula. 

1. The area of a parallelogram determined by vectors \(\vec{v}\) and \(\vec{w}\) is given by the magnitude of the cross product, \(|\vec{v} \times \vec{w}|\).
2. We are given that the area of the parallelogram with sides \(\vec{a}\) and \(\vec{b}\) is 15, which implies: \(|\vec{a} \times \vec{b}| = 15\).
3. We need to find \(| (3\vec{a} + 2\vec{b}) \times (\vec{a} + 3\vec{b}) |\).
4. Using the distributive property of the cross product, we have: \((3\vec{a} + 2\vec{b}) \times (\vec{a} + 3\vec{b}) = 3\vec{a} \times \vec{a} + 3\vec{a} \times 3\vec{b} + 2\vec{b} \times \vec{a} + 2\vec{b} \times 3\vec{b}\).
5. Since the cross product of any vector with itself is zero (\(\vec{v} \times \vec{v} = 0\)), we simplify: \(= 9(\vec{a} \times \vec{b}) + 2(\vec{b} \times \vec{a})\).
6. Recall that \(\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})\), hence: \(= 9(\vec{a} \times \vec{b}) - 2(\vec{a} \times \vec{b}) = 7(\vec{a} \times \vec{b})\).
7. Thus, the area of the new parallelogram is the absolute value: \(|7(\vec{a} \times \vec{b})| = 7 |\vec{a} \times \vec{b}| = 7 \times 15 = 105\).

Therefore, the area of the parallelogram with sides \(3\vec{a} + 2\vec{b}\) and \(\vec{a} + 3\vec{b}\) is 105 s units.