The resultant of P and Q is perpendicular to P- What is the angle between - P and QA- cos -1 p - Q B- cos -1- p - Q C- sin -1 p - Q D- sin -1-p-Q

Let R be the resultant vector and α be the angle between nbsp;R and nbsp;P Using formula, tanα = b sinθa + b cosθ tan 90 = Q sin θP + Q cos θ ⇒ P + Q cos θ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Vector Component

The resultant...

Question

The resultant of $\to P$ and $\to Q$ is perpendicular to $\to P$ . What is the angle between $\to P$ and $\to Q$

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

{sin}^{- 1} (\frac{P}{Q})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{sin}^{- 1} (- \frac{P}{Q})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B ${cos}^{- 1} (- \frac{P}{Q})$
Let $\to R$ be the resultant vector and $α$ be the angle between $\to R$ and $\to P$
Using formula,
$tan α = \frac{b sin θ}{a + b cos θ}$
$tan 90 = \frac{Q sin θ}{P + Q cos θ}$
$\Rightarrow P + Q cos θ = 0$
$cos θ = \frac{- P}{Q}$
$θ = {cos}^{- 1} (\frac{- P}{Q})$

Suggest Corrections

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

The correct option is B cos−1(−PQ)Let →R be the resultant vector and α be the angle between →R and →P Using formula, tanα=bsinθa+bcosθ tan 90=Q sin θP+Q cos θ ⇒P+Q cos θ=0 cosθ=−PQ θ=cos−1(−PQ)

The resultant of $\to P$ and $\to Q$ is perpendicular to $\to P$ . What is the angle between $\to P$ and $\to Q$