The Question is from neet-physics motion-in-a-plane , and it was created by the author ABDUL JILANI BASHA

Question

$\mathop {{\rm{ }}A}\limits^ -   = \mathop {{\rm{ }}B}\limits^ -   + \mathop {{\rm{ }}C}\limits^ -  $ and the magnitudes of $\mathop {{\rm{ }}A}\limits^ -  ,\mathop {{\rm{ }}B}\limits^ -  ,\mathop {{\rm{ }}C}\limits^ -  $ are 5, 4 and 3 units respectively, the angle between $\mathop {{\rm{ }}A}\limits^ -  \mathop {{\rm{ }}C}\limits^ -  $ is:

ABDUL JILANI BASHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
**Primary Concept**: Vector addition and dot product to find the angle between vectors
**Related Topics**: Vector magnitudes, cosine rule
**Prerequisites**: Basic vector algebra, dot product formula
## SOLUTION PATHWAY
**Given Information**:
   - $\mathop {{m{ }}A}\limits^ -  = \mathop {{m{ }}B}\limits^ -  + \mathop {{m{ }}C}\limits^ -  $
   - $|\mathop {{m{ }}A}\limits^ -  | = 5$
   - $|\mathop {{m{ }}B}\limits^ -  | = 4$
   - $|\mathop {{m{ }}C}\limits^ -  | = 3$

**Method & Steps**:
   1. Isolate vector C: $\mathop {{m{ }}C}\limits^ -  = \mathop {{m{ }}A}\limits^ -  - \mathop {{m{ }}B}\limits^ -  $
      Reasoning: This allows us to work with A and C directly
    2. Take the dot product of C with itself: $\mathop {{m{ }}C}\limits^ -  \cdot \mathop {{m{ }}C}\limits^ -  = (\mathop {{m{ }}A}\limits^ -  - \mathop {{m{ }}B}\limits^ -  ) \cdot (\mathop {{m{ }}A}\limits^ -  - \mathop {{m{ }}B}\limits^ -  )$
      Reasoning: The dot product of a vector with itself is the square of its magnitude, and expanding the right side will give us terms we can use
      Formula/Rule: $\mathop {{m{ }}V}\limits^ -  \cdot \mathop {{m{ }}V}\limits^ -  = |\mathop {{m{ }}V}\limits^ -  |^2$
    3. Expand the dot product: $|\mathop {{m{ }}C}\limits^ -  |^2 = |\mathop {{m{ }}A}\limits^ -  |^2 + |\mathop {{m{ }}B}\limits^ -  |^2 - 2\mathop {{m{ }}A}\limits^ -  \cdot \mathop {{m{ }}B}\limits^ -  $
      Working: Using the distributive property of dot product
    4. Calculate the dot product of A and B : $3^2 = 5^2 + 4^2 - 2\mathop {{m{ }}A}\limits^ -  \cdot \mathop {{m{ }}B}\limits^ -  $
     Working: Substituting the values $9 = 25 + 16 - 2\mathop {{m{ }}A}\limits^ -  \cdot \mathop {{m{ }}B}\limits^ -  $
      Key Point: Simplify this equation to find  $\mathop {{m{ }}A}\limits^ -  \cdot \mathop {{m{ }}B}\limits^ -  = 16$
   5. We know that $\mathop {{m{ }}A}\limits^ -  \cdot \mathop {{m{ }}C}\limits^ -  = |\mathop {{m{ }}A}\limits^ -  ||\mathop {{m{ }}C}\limits^ -  |\cos	heta$ and  $\mathop {{m{ }}C}\limits^ -  = \mathop {{m{ }}A}\limits^ -  - \mathop {{m{ }}B}\limits^ -  $
   6. Substituting and expanding: $\mathop {{m{ }}A}\limits^ -  \cdot (\mathop {{m{ }}A}\limits^ -  - \mathop {{m{ }}B}\limits^ -  ) = |\mathop {{m{ }}A}\limits^ -  |^2 - \mathop {{m{ }}A}\limits^ -  \cdot \mathop {{m{ }}B}\limits^ -  = 25 - 16 = 9$
   7. Now, $ \mathop {{m{ }}A}\limits^ -  \cdot \mathop {{m{ }}C}\limits^ - = |\mathop {{m{ }}A}\limits^ -  ||\mathop {{m{ }}C}\limits^ -  |\cos	heta = 9$
   8. Substitute known values: $5 * 3 * \cos	heta = 9$
   9. Calculate  $\cos	heta = \frac{9}{15} = \frac{3}{5}$
   10. Find the angle: $	heta = \cos^{-1}\left(\frac{3}{5}ight)$

**Result**: $	heta = \cos^{-1}\left(\frac{3}{5}ight)$

## OPTION ANALYSIS
[A] ✓:  The angle between A and C is $\cos^{-1}\left(\frac{3}{5}ight)$, which matches the calculated result.
[B] :x:  This would be the angle if B was the resultant vector, not C.
[C] :x: The vectors are not orthogonal
[D] :x: This option provides the sine inverse, not cosine and also uses incorrect parameters
## LEARNING AIDS
**Common Mistakes**:
   1. Misusing the dot product property: Ensure that the dot product is expanded correctly.
   2. Incorrectly isolating the required vector: Make sure you manipulate the equations correctly.
**Quick Tips**:
   1. Use the dot product to find angles between vectors.
   2. Draw a diagram to visualize the vectors and their relationship to each other.

ABDUL JILANI BASHA · Answer

The correct Option is "${{\mathop{m Cos}
olimits} ^{ - 1}}\left( {\frac{3}{5}} ight)$"

ABDUL JILANI BASHA · Answer

${{\mathop{m Cos}
olimits} ^{ - 1}}\left( {\frac{4}{5}} ight)$

ABDUL JILANI BASHA · Answer

$\frac{\pi }{2}$

ABDUL JILANI BASHA · Answer

${\sin ^{ - 1}}\left( {\frac{3}{4}} \right)$

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