The resultant of A- and B- is perpendicular to A- What is the angle between A- and B-

The correct option is C cos^ 1( AB) From the figure #10; #10; A⃗=B⃗cos( 180^o θ #160; ) #10; #10; θ =cos^ 1( A⃗B⃗ #10;) #160; ...

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Standard XII

Physics

Vector Component

The resultant...

Question

The resultant of $\to A$ and $\to B$ is perpendicular to $\to A$ . What is the angle between $\to A$ and $\to B$ ?

{cos}^{- 1} (\frac{A}{B})

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{cos}^{- 1} (- \frac{A}{B})

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{sin}^{- 1} (\frac{A}{B})

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{sin}^{- 1} (- \frac{A}{B})

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Solution

The correct option is C ${cos}^{- 1} (- \frac{A}{B})$
From the figure

$\to A = \to B cos (180^{o} - θ)$

$θ = {cos}^{- 1} (\frac{- \to A}{\to B})$

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The resultant of →A and →B is perpendicular to →A. What is the angle between →A and →B?

The correct option is C cos−1(−AB)From the figure →A=→Bcos(180o−θ) θ=cos−1(−→A→B)

The resultant of $\to A$ and $\to B$ is perpendicular to $\to A$ . What is the angle between $\to A$ and $\to B$ ?

The correct option is C ${cos}^{- 1} (- \frac{A}{B})$
From the figure

$\to A = \to B cos (180^{o} - θ)$

$θ = {cos}^{- 1} (\frac{- \to A}{\to B})$