The resultant of two vectors A and B is perpendicular to A and its magnitude is half that of B. The angle between vectors A and B is) _______ ^ .

Question

Accepted Answer

Correct answer: 150. (150) To solve this problem, we'll analyze the conditions given about the vectors A and B, and their resultant R. Given: (i) R=A+B is perpendicular to A. (ii) The magnitude of R is half that of B : |R |= 12 |B | . Approach: We know that if R is perpendicular to A, then their dot product is zero: R A=0 Substitute R=A+B : (A+B ) A=0 ⇒A A+B A=0 ⇒ |A |^2+ |A | |B | cos ⁡ =0 ⇒ cos ⁡ =- |A |^2 |A | |B | ⇒ cos ⁡ =- |A | |B | Next, we use the second condition involving magnitudes: |R |= |A+B |= 12 |B | ⇒ |A |^2+ |B |^2+2 |A | |B | cos ⁡ = 12 |B | ⇒ |A |^2+ |B |^2-2 |A | |B | |A | |B |= 14 |B |^2 ⇒ |A |^2+ |B |^2-2 |A |^2= 14 |B |^2 ⇒ |B |^2- |A |^2= 14 |B |^2 ⇒ 34 |B |^2= |A |^2 ⇒ |A |= 32 |B | Finally, substitute this back into the cosine formula: cos =- 32 (B ) (B ) cos ⁡ =- 32 This value of cos ⁡ corresponds to an angle of 150^ because cos ⁡150^ =- 32. The angle between A and B is 150^ .

$The resultant of two vectors A and B is perpendicular to A and its magnitude is half that of B . The angle between vectors A and B is) _______^{\circ} .$

Solution

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