Question

29.
Two vectors A and B have equal magnitudes. The magnitude of (A+B) is 'n times the magnitude of
(A-B). The angle between A and B is:
(JEE (Main) 2019-Online)
(A) sin1
n² -
(B) cos-1
n- +1
(C) cos-1
(D) sir? n+1]
+1
n
-1​

Answer 1

Given :-

(|A+B|) = n(|A-B|)

Squaring both sides.

(A² + B² + 2AB CosØ) = n²(A² + B² - 2AB CosØ)

It is given that the magnitudes of A and B is equal. Taking,

|A| = |B|

(2A² + 2A²CosØ) = n²(2A² - 2A²CosØ)

2A²(1 + CosØ) = 2n²A²(1 - CosØ)

1 + CosØ = n²(1 - CosØ)

CosØ + n²CosØ = n² - 1

CosØ(1 + n²) = n² - 1

Ø = Cos-¹[n²-1/n²+1]

Hence,

The angle between A and B = Ø = Cos-¹[n²-1/n²+1] .

Answer 2

AppropriateQuestion: Two vectors A and B have equal magnitudes. The magnitude of (A+B) is 'n times the magnitude of

(A-B). The angle between A and B is:

_______________________________

• To find the correct answer.

The Answer is $\sf {cos}^{ - 1} \: \big (\frac{ n^{2} - 1}{ {n}^{2 - 1} } )$

Kindly, See the attached for more clarification.