The Question is from neet-physics motion-in-a-plane , and it was created by the author Mr. ANIL KUMAR.G

Question

The area of a parallelogram whose adjacent sides are given by vectors $\vec A = \hat i - 2\hat j + 3\hat k$ and $\vec B = 4\hat i + 5\hat j$  is

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: D
## CONCEPT FRAMEWORK
	**Primary Concept**: Area of a parallelogram formed by two vectors.
	**Related Topics**: Vector cross product, magnitude of a vector.
	**Prerequisites**: Knowledge of vector representation, cross product calculation, and magnitude calculation of a vector.
## SOLUTION PATHWAY
**Given Information**:
	* Vector A = \hat i - 2\hat j + 3\hat k
	* Vector B = 4\hat i + 5\hat j
	* Area of parallelogram = |\vec A x \vec B|
**Method & Steps**:
	1. Calculate the cross product \vec A x \vec B.
		Reasoning: The area of a parallelogram formed by two adjacent vectors is equal to the magnitude of their cross product.
		Formula/Rule:  $\vec A 	imes \vec B = \begin{vmatrix} \hat i & \hat j & \hat k \ a_x & a_y & a_z \ b_x & b_y & b_z \end{vmatrix}$
	2.  Calculate the cross product: $\vec A 	imes \vec B = \begin{vmatrix} \hat i & \hat j & \hat k \ 1 & -2 & 3 \ 4 & 5 & 0 \end{vmatrix} = \hat i((-2)(0) - (3)(5)) - \hat j((1)(0) - (3)(4)) + \hat k((1)(5) - (-2)(4)) = -15\hat i + 12\hat j + 13\hat k$
		Working: \vec A x \vec B = -15\hat i + 12\hat j + 13\hat k
		Key Point: Correct cross product calculation is crucial.
	3. Calculate the magnitude of the resulting vector.
		Reasoning: The magnitude of the cross product gives the area of the parallelogram.
		Formula/Rule:  |\vec V| = $\sqrt{V_x^2 + V_y^2 + V_z^2}$
	4.  Magnitude Calculation: |\vec A x \vec B| = $\sqrt{(-15)^2 + (12)^2 + (13)^2} = \sqrt{225 + 144 + 169} = \sqrt{538}$
		Result: |\vec A x \vec B| = $\sqrt{538}$
		Verification: The calculation is correct and the magnitude is the area.
## OPTION ANALYSIS
[A] :x: 10 is not the correct area.
[B] :x: 40 is not the correct area.
[C] :x: $\sqrt{88}$ is not the correct area.
[D] ✓: $\sqrt{538}$ is the correct area obtained by calculating the magnitude of the cross product of the vectors.
## LEARNING AIDS
	**Common Mistakes**:
		1. **Incorrect Cross Product Calculation**: Ensure the determinant is calculated correctly with proper signs.
		2. **Incorrect Magnitude Calculation**: Double-check squaring and summing of components, and avoid mistakes in square root calculations.
	**Quick Tips**:
		1. **Visualizing Vectors**: Try visualizing the vectors to understand the area concept.
		2. **Determinant Formula**: Practice calculating determinants to avoid errors.

Mr. ANIL KUMAR.G · Answer

10

Mr. ANIL KUMAR.G · Answer

40

Mr. ANIL KUMAR.G · Answer

$\sqrt {88} $

Mr. ANIL KUMAR.G · Answer

The correct Option is "$\sqrt {538} $"

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