The angle between vector ( vecA) and ( vecA- vecB) is :

Question

The angle between vector ( vecA) and ( vecA- vecB) is : (A)  tan -1((-( B /2)/ A - B (√3)2)) (B)  tan -1(( A /0.7 B )) (C)  tan -1((√3 B /2 A - B )) (D)  tan -1(( B cos θ/ A - B sin θ))

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/06ec25c26eec09d432c63976da16982b-.png" /> Angle between vec A and vec B , θ=60° Angle betwenn vecA and vecA- vecB tan α=( B sin θ/ A - B cos θ) =(B √(3/2)/A-B × (1)2 2) tan α=(√3 B/2 A-B)

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

$tan^{- 1} (\frac{- \frac{B}{2}}{A - B \frac{3}{2}})$

$tan^{- 1} (\frac{A}{0.7 B})$

$tan^{- 1} (\frac{3 B}{2 A - B})$

$tan^{- 1} (\frac{B c o s θ}{A - B s i n θ})$

Angle between $A$ and $B, θ = 6 0^{\circ}$
Angle betwenn $A$ and $A - B$
$tan α = \frac{B s i n θ}{A - B c o s θ}$
$= \frac{B \frac{3}{2}}{A - B \times \frac{1}{2} 2}$
$tan α = \frac{3 B}{2 A - B}$

Q. The angle between vector (A) and (A−B) is :

tan−1(A−B23​​−2B​​)

tan−1(0.7BA​)

tan−1(2A−B3​B​)

tan−1(A−BsinθBcosθ​)