The vectors A- and B- are such that A- - B- - C- and A2 - B2 - C2- Angle - between positive directions of A- and B- is

The correct option is A π2 Given, A⃗+B⃗=C⃗ and A^2+B^2=C^2 Then (A+B).(A+B)=C.C A.A+A.B+B.A+B.B=C.C =A^2+2A.B+B^2=C^2 =2A.B=0 ...

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Byju's Answer

Standard XII

Physics

Work Done as Dot Product

The vectors ...

Question

The vectors $\to A$ and $\to B$ are such that $\to A + \to B = \to C$ and $A^{2} + B^{2} = C^{2}$ . Angle $θ$ between positive directions of $\to A$ and $\to B$ is

\frac{π}{2}

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0

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π

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\frac{2 π}{3}

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Solution

The correct option is A $\frac{π}{2}$
Given,

$\to A + \to B = \to C$ and $A^{2} + B^{2} = C^{2}$

Then

$(A + B) . (A + B) = C . C$

$⟹ A . A + A . B + B . A + B . B = C . C$

$= A^{2} + 2 A . B + B^{2} = C^{2}$

$= 2 A . B = 0$

$= A . B = 0$

$⟹ | A | | B | c o s θ = 0$

$⟹ c o s θ = 0$

So, Angle $θ = \frac{π}{2}$

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The vectors →A and →B are such that →A+→B=→C and A2+B2=C2. Angle θ between positive directions of →A and →B is

The correct option is A π2Given,→A+→B=→C and A2+B2=C2Then(A+B).(A+B)=C.C⟹A.A+A.B+B.A+B.B=C.C=A2+2A.B+B2=C2=2A.B=0=A.B=0⟹|A||B|cosθ=0⟹cosθ=0So, Angle θ=π2

The vectors $\to A$ and $\to B$ are such that $\to A + \to B = \to C$ and $A^{2} + B^{2} = C^{2}$ . Angle $θ$ between positive directions of $\to A$ and $\to B$ is