Question

If area of the parallelogram with $\vec{a}$ and $\vec{b}$ as two adjacent sides is 20 square units, then the area of the parallelogram having $3\vec{a}+\vec{b}$ and $2\vec{a}+3\vec{b}$ as two adjacent sides in square units is

• A. 105
• B. 120
• C. 75
• D. 140

Hint:

You can use a determinant shortcut for vectors formed by linear combinations: the new area is simply the determinant of the scalar coefficients multiplied by the original area.
$\begin{vmatrix} 3 & 1 \\ 2 & 3 \end{vmatrix} \times 20 = (9 - 2) \times 20 = 7 \times 20 = 140$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given the area of a base parallelogram formed by vectors $\vec{a}$ and $\vec{b}$. We need to find the area of a new parallelogram formed by linear combinations of these vectors.

Step 2: Key Formula or Approach:
The area of a parallelogram with adjacent sides $\vec{u}$ and $\vec{v}$ is the magnitude of their cross product: $\text{Area} = |\vec{u} \times \vec{v}|$.
Properties of the cross product used here:
1. $\vec{x} \times \vec{x} = \vec{0}$
2. $\vec{x} \times \vec{y} = -(\vec{y} \times \vec{x})$

Step 3: Detailed Explanation:
The area of the original parallelogram is given as:
$$|\vec{a} \times \vec{b}| = 20$$ Let the adjacent sides of the new parallelogram be $\vec{A} = 3\vec{a} + \vec{b}$ and $\vec{B} = 2\vec{a} + 3\vec{b}$.
The new area is $|\vec{A} \times \vec{B}|$. First, expand the cross product:
$$\vec{A} \times \vec{B} = (3\vec{a} + \vec{b}) \times (2\vec{a} + 3\vec{b})$$ Use the distributive property:
$$= 3\vec{a} \times (2\vec{a} + 3\vec{b}) + \vec{b} \times (2\vec{a} + 3\vec{b})$$ $$= 6(\vec{a} \times \vec{a}) + 9(\vec{a} \times \vec{b}) + 2(\vec{b} \times \vec{a}) + 3(\vec{b} \times \vec{b})$$ Since the cross product of any vector with itself is zero ($\vec{a} \times \vec{a} = \vec{0}$ and $\vec{b} \times \vec{b} = \vec{0}$):
$$= \vec{0} + 9(\vec{a} \times \vec{b}) + 2(\vec{b} \times \vec{a}) + \vec{0}$$ Since $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$:
$$= 9(\vec{a} \times \vec{b}) - 2(\vec{a} \times \vec{b})$$ $$= 7(\vec{a} \times \vec{b})$$ Now, find the magnitude to get the area:
$$\text{New Area} = |7(\vec{a} \times \vec{b})| = 7|\vec{a} \times \vec{b}|$$ Substitute the given value of the original area ($20$):
$$\text{New Area} = 7 \times 20 = 140$$

Step 4: Final Answer:
The area of the new parallelogram is 140 square units, matching option (D).