Question

Let \(\vec{a} = \hat{i} + 2\hat{j}\) and \(\vec{b} = 2\hat{i} + \hat{j}\). Is \(|\vec{a}| = |\vec{b}|\)? Are the vectors \(\vec{a}\) and \(\vec{b}\) equal?

Hint:

Having the same magnitude does not imply that two vectors are equal. Equal vectors must have both the same magnitude and the same direction. Different vectors can have the same length but point in different directions.

Answer 1

Step 1: Understanding the Concept:
This question checks the understanding of two different properties of vectors: magnitude and equality.
Magnitude of a vector: The length of the vector. For \(\vec{v} = x\hat{i} + y\hat{j}\), the magnitude is \(|\vec{v}| = \sqrt{x^2+y^2}\).
Equality of vectors: Two vectors are equal if and only if all their corresponding components are identical.
Step 2: Key Formula or Approach:
1. Calculate \(|\vec{a}|\) and \(|\vec{b}|\) using the magnitude formula and compare them.
2. Compare the corresponding components of \(\vec{a}\) and \(\vec{b}\) to check for equality.
Step 3: Detailed Explanation or Calculation:
Given vectors: \[ \vec{a} = 1\hat{i} + 2\hat{j} \] \[ \vec{b} = 2\hat{i} + 1\hat{j} \] 1. Check if Magnitudes are Equal:
Calculate the magnitude of \(\vec{a}\): \[ |\vec{a}| = \sqrt{(1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5} \] Calculate the magnitude of \(\vec{b}\): \[ |\vec{b}| = \sqrt{(2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5} \] Since both magnitudes are equal to \(\sqrt{5}\), we can conclude that \(|\vec{a}| = |\vec{b}|\).
2. Check if Vectors are Equal:
For two vectors to be equal, their corresponding components must be equal.
Vector \(\vec{a}\) can be written as \((1, 2)\).
Vector \(\vec{b}\) can be written as \((2, 1)\).
Comparing the \(\hat{i}\) components: \(1 \neq 2\).
Comparing the \(\hat{j}\) components: \(2 \neq 1\).
Since the components are not identical, the vectors \(\vec{a}\) and \(\vec{b}\) are not equal.
Step 4: Final Answer:
Yes, the magnitudes are equal: \(|\vec{a}| = |\vec{b}| = \sqrt{5}\).
No, the vectors are not equal because their corresponding components are different.