SATHEE: Kinematics Question 22

Kinematics Question 22

Question: The resultant of $\vec{P}$ and $\vec{Q}$ perpendicular to $\vec{P}$ . What is the angle between $\vec{P}$ and $\vec{Q}$

Options:

A) $\cos^{- 1} (P / Q)$

B) $\cos^{- 1} (- P / Q)$

C) $\sin^{- 1} (P / Q)$

D) $\sin^{- 1} (- P / Q)$

Show Answer

Answer:

Correct Answer: B

Solution:

therefore $\tan 90^{\circ} = \frac{Q \sin θ}{P + Q \cos θ}$

therefore $P + Q \cos θ = 0$

$\cos θ = \frac{- P}{Q}$ $θ = \cos^{- 1} (\frac{- P}{Q})$

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Kinematics Question 22

Question: The resultant of P→ and Q→ perpendicular to P→ . What is the angle between P→ and Q→

Options:

Answer:

Solution:

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Question: The resultant of $\vec{P}$ and $\vec{Q}$ perpendicular to $\vec{P}$ . What is the angle between $\vec{P}$ and $\vec{Q}$