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

Question

Given that ${\rm{\vec A}} + {\rm{\vec B}} + {\rm{\vec C}} = 0$. Out of three vectors, two are equal in magnitude and the magnitude of third vector is $\sqrt 2 $ times that of either of the two having equal magnitude. Then the angles between vectors are given by

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Vector Addition and Properties
	**Related Topics**: Law of Cosines, Dot Product
	**Prerequisites**: Basic vector algebra, trigonometric identities
## SOLUTION PATHWAY
**Given Information**:
	* ${\vec{A}} + {\vec{B}} + {\vec{C}} = 0$
	* Let $|{\vec{A}}| = |{\vec{B}}| = a$ and $|{\vec{C}}| = a\sqrt{2}$
**Method & Steps**:
	1. Rearrange the vector equation: ${\vec{A}} + {\vec{B}} = -{\vec{C}}$
		Reasoning: To find the angles, we'll use the magnitudes and the relationship between the vectors.
		Formula/Rule: Vector addition property
	2. Square both sides of the equation: $|{\vec{A}} + {\vec{B}}|^2 = |-{\vec{C}}|^2$
		Working: $({\vec{A}} + {\vec{B}}) \cdot ({\vec{A}} + {\vec{B}}) = |{\vec{C}}|^2$
		Key Point: Using dot product to find magnitudes
	3. Expand and simplify: $|{\vec{A}}|^2 + |{\vec{B}}|^2 + 2|{\vec{A}}||{\vec{B}}|\cos{	heta_{AB}} = |{\vec{C}}|^2$
		Working: $a^2 + a^2 + 2a^2\cos{	heta_{AB}} = (a\sqrt{2})^2$ 
		Key Point: $	heta_{AB}$ is the angle between vectors A and B.
	4. Solve for $\cos{	heta_{AB}}$: $2a^2 + 2a^2\cos{	heta_{AB}} = 2a^2$ which implies $2a^2\cos{	heta_{AB}} = 0$ and hence $\cos{	heta_{AB}} = 0$.
		Result: Therefore, $	heta_{AB} = 90^\circ$
	5. Similarly, rearrange: ${\vec{A}} + {\vec{C}} = -{\vec{B}}$. Then $|{\vec{A}} + {\vec{C}}|^2 = |{\vec{B}}|^2$
		Reasoning: To find the angle between A and C
	6. Expand: $|{\vec{A}}|^2 + |{\vec{C}}|^2 + 2|{\vec{A}}||{\vec{C}}|\cos{	heta_{AC}} = |{\vec{B}}|^2$
		Working: $a^2 + 2a^2 + 2a(a\sqrt{2})\cos{	heta_{AC}} = a^2$ which means $3a^2 + 2a^2\sqrt{2}\cos{	heta_{AC}} = a^2$
	7. Solve for $\cos{	heta_{AC}}$: $2a^2\sqrt{2}\cos{	heta_{AC}} = -2a^2$, hence $\cos{	heta_{AC}} = -\frac{1}{\sqrt{2}}$
		Result: Thus $	heta_{AC} = 135^\circ$
	8. Similarly, the angle between B and C will be $	heta_{BC} = 135^\circ$ due to symmetry.
		Verification: The angles match option B.
## OPTION ANALYSIS
[A] :x: Incorrect angles
[B] ✓: Correct angles derived above
[C] :x: Incorrect angles
[D] :x: Incorrect angles
## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Forgetting to square both sides when dealing with magnitudes. [How to avoid]: Always remember to square when working with magnitudes in vector equations.
		2.[Error 2]: Incorrectly applying the law of cosines. [How to avoid]: Ensure correct placement of vectors and angles in the formula.
	**Quick Tips**:
		1.[Helpful shortcut]: Visualizing the vector addition can help understand the angle relationships.
		2.[Memory aid]: Remember that when the sum of two vectors equals the negative of the third, they form a closed triangle.

Mr. ANIL KUMAR.G · Answer

$45^\circ ,45^\circ ,90^\circ $

Mr. ANIL KUMAR.G · Answer

The correct Option is "$90^\circ ,135^\circ ,135^\circ $"

Mr. ANIL KUMAR.G · Answer

$30^\circ ,60^\circ ,90^\circ $

Mr. ANIL KUMAR.G · Answer

$45^\circ ,60^\circ ,90^\circ $

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