SATHEE: Kinematics Question 19

Kinematics Question 19

Question: The resultant of two vectors A and b is perpendicular to the vector A and it’s magnitude is equal to half the magnitude of vector B. The angle between A and b is

Options:

A) $120^{\circ}$

B) $150^{\circ}$

C) $135^{\circ}$

D) None of these

Show Answer

Answer:

Correct Answer: B

Solution:

$\frac{B}{2} = \sqrt{A^{2} + B^{2} + 2 A B \cos θ}$ -(i)

$\tan 90^{\circ} = \frac{B \sin θ}{A + B \cos θ} \Rightarrow A + B \cos θ = 0$ $\cos θ = - \frac{A}{B}$

Hence, from (i) $\frac{B^{2}}{4} = A^{2} + B^{2} - 2 A^{2} \Rightarrow A = \sqrt{3} \frac{B}{2}$

therefore $\cos θ = - \frac{A}{B} = - \frac{\sqrt{3}}{2}$ $θ = 150^{\circ}$

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Kinematics Question 19

Question: The resultant of two vectors A and b is perpendicular to the vector A and it’s magnitude is equal to half the magnitude of vector B. The angle between A and b is

Options:

Answer:

Solution:

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