The Question is from neet-chemistry , and it was created by the author A.RAMA NAGESWARARAO

Question

If $\vec{A} = 2\hat{i} + 3\hat{j}$ and $\vec{B} = \hat{i} - 2\hat{j}$, a vector $\vec{C}$ which is perpendicular to both $\vec{A}$ and $\vec{B}$ and has a magnitude of 5 units can be:

A.RAMA NAGESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Cross Product of Vectors
	**Related Topics**: Vector Algebra, Perpendicular Vectors, Magnitude of a Vector
	**Prerequisites**: Understanding of vectors, cross product, and magnitude calculations.
## SOLUTION PATHWAY
**Given Information**:
	$\vec{A} = 2\hat{i} + 3\hat{j}$
	$\vec{B} = \hat{i} - 2\hat{j}$
	$|\vec{C}| = 5$
	$\vec{C}$ is perpendicular to both $\vec{A}$ and $\vec{B}$
**Method & Steps**:
	1. **Calculate the cross product of** $\vec{A}$ **and** $\vec{B}$:
		Reasoning: The cross product of two vectors results in a vector that is perpendicular to both.
		Formula/Rule: $\vec{A} 	imes \vec{B} = (A_x\hat{i} + A_y\hat{j}) 	imes (B_x\hat{i} + B_y\hat{j}) = (A_xB_y - A_yB_x)\hat{k}$
		Working: $\vec{A} 	imes \vec{B} = (2(-2) - 3(1))\hat{k} = (-4 - 3)\hat{k} = -7\hat{k}$
	2. **Find a unit vector in the direction of** $\vec{A} 	imes \vec{B}$:
		Working: $\hat{n} = \frac{\vec{A} 	imes \vec{B}}{|\vec{A} 	imes \vec{B}|} = \frac{-7\hat{k}}{\sqrt{(-7)^2}} = \frac{-7\hat{k}}{7} = -\hat{k}$
		Key Point:  A unit vector has a magnitude of 1.
	3. **Multiply the unit vector by the desired magnitude**:
		Result: $\vec{C} = \pm 5\hat{k}$ (Since the magnitude is 5, and the direction can be either positive or negative k)
		Verification: The magnitude of $5\hat{k}$ and $-5\hat{k}$ is 5. Both are perpendicular to $\vec{A}$ and $\vec{B}$.
## OPTION ANALYSIS
[A] :x: $7\hat{i} - \hat{k}$ is not perpendicular to both $\vec{A}$ and $\vec{B}$.
[B] ✓: $5\hat{k}$ or $-5\hat{k}$ are perpendicular to both $\vec{A}$ and $\vec{B}$ and have a magnitude of 5.
[C] :x: $-7\hat{k}$ has a magnitude of 7, not 5.
[D] :x: $5\hat{i} + 5\hat{j}$ is not perpendicular to both $\vec{A}$ and $\vec{B}$.
## LEARNING AIDS
	**Common Mistakes**:
		1. Forgetting to consider both positive and negative directions of the cross product: Remember that a vector perpendicular to two given vectors can point in two opposite directions.
		2. Confusing dot product and cross product: Dot product gives a scalar, while cross product gives a vector.
	**Quick Tips**:
		1. Visualize the vectors in 3D space to understand the direction of the cross product.
		2. Remember the right-hand rule for determining the direction of the cross product.

A.RAMA NAGESWARARAO · Answer

7i - k

A.RAMA NAGESWARARAO · Answer

The correct Option is "5k or -5k"

A.RAMA NAGESWARARAO · Answer

-7k

A.RAMA NAGESWARARAO · Answer

5i + 5j

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