Question:medium

Vectors of magnitude \(3\) and \(4\) are perpendicular. Resultant?

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For perpendicular vectors, use Pythagoras theorem: \[ R=\sqrt{A^2+B^2} \] The pair \(3,4,5\) is a common right-triangle result.
Updated On: Jun 3, 2026
  • \(5\)
  • \(7\)
  • \(1\)
  • \(12\)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
When two vectors are added, the magnitude of the resultant vector depends on the magnitudes of the original vectors and the angle between them.
According to the parallelogram law of vector addition, the magnitude of the resultant \( R \) is given by a formula involving a square root.
When the vectors are perpendicular (\( \theta = 90^\circ \)), the formula simplifies significantly.
Step 2: Key Formula or Approach:
The general formula for the resultant of two vectors \( \vec{A} \) and \( \vec{B} \) is:
\[ R = \sqrt{A^{2} + B^{2} + 2AB \cos \theta} \]
For perpendicular vectors, \( \theta = 90^\circ \). Since \( \cos 90^\circ = 0 \), the formula reduces to the Pythagorean theorem:
\[ R = \sqrt{A^2 + B^2} \]
Step 3: Detailed Explanation:
1. Identify the magnitudes:
- Magnitude of vector 1 (\( A \)) = 3 units.
- Magnitude of vector 2 (\( B \)) = 4 units.
2. Note that they are perpendicular, meaning the angle between them is \( 90^\circ \).
3. Apply the Pythagorean theorem:
\[ R = \sqrt{3^{2} + 4^{2}} \]
4. Calculate the squares:
\[ R = \sqrt{9 + 16} \]
5. Sum the values:
\[ R = \sqrt{25} \]
6. Take the square root:
\[ R = 5 \text{ units} \]
Step 4: Final Answer:
The magnitude of the resultant vector is 5.
This is a standard "Pythagorean triple" result.
Correct option is (A).
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