Question

If the area of the parallelogram with \(\vec{a}\) and \(\vec{b}\) as two adjacent sides is 15 sq.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 sq.units is

• A. 45
• B. 75
• C. 105
• D. 120

Answer 1

The Correct Option is C

The area of a parallelogram formed by two vectors \( \vec{a} \) and \( \vec{b} \) is given by the magnitude of the cross product of these vectors: \[ \text{Area of parallelogram} = |\vec{a} \times \vec{b}| \] We are given that the area of the parallelogram with sides \( \vec{a} \) and \( \vec{b} \) is 15 square units, so: \[ |\vec{a} \times \vec{b}| = 15 \] Now, we need to find the area of the parallelogram formed by the vectors \( 3\vec{a} + 2\vec{b} \) and \( \vec{a} + 3\vec{b} \). The area of the new parallelogram is given by the magnitude of the cross product of these two vectors: \[ \text{Area} = |(3\vec{a} + 2\vec{b}) \times (\vec{a} + 3\vec{b})| \] We expand the cross product: \[ (3\vec{a} + 2\vec{b}) \times (\vec{a} + 3\vec{b}) = 3\vec{a} \times \vec{a} + 3\vec{a} \times 3\vec{b} + 2\vec{b} \times \vec{a} + 2\vec{b} \times 3\vec{b} \] Using the property that \( \vec{a} \times \vec{a} = \vec{0} \) and \( \vec{b} \times \vec{b} = \vec{0} \), we get: \[ = 9(\vec{a} \times \vec{b}) + 2(\vec{b} \times \vec{a}) \] Since \( \vec{b} \times \vec{a} = -(\vec{a} \times \vec{b}) \), we get: \[ = 9(\vec{a} \times \vec{b}) - 2(\vec{a} \times \vec{b}) = 7(\vec{a} \times \vec{b}) \] Thus, the magnitude of this cross product is: \[ \text{Area} = |7(\vec{a} \times \vec{b})| = 7|\vec{a} \times \vec{b}| = 7 \times 15 = 105 \]

So, the correct answer is (C) : 105.