The magnitude of vectors A - B and C are respectively 12-5 and 13 units and A - B - C- then the angle between A and B isA- 0B- TC- - - 2D- - - 4

The correct option is C π/2Given, C=A+B C.C=(A+B).(A+B) C^2=A^2+B^2+2AB cosθ 13^2=12^2+5^2+2(12)(5)cosθ cosθ=0 ⇒θ=π2 ...

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Byju's Answer

Standard XII

Physics

Unit Vectors

The magnitude...

Question

The magnitude of vectors $\to A, \to B$ and $\to C$ are respectively $12, 5$ and $13$ units and $\to A + \to B = \to C$ , then the angle between $\to A$ and $\to B$ is

0

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π

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π / 2

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π / 4

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Solution

The correct option is C $π / 2$
Given, $\to C = \to A + \to B$
$\to C . \to C = (\to A + \to B) . (\to A + \to B)$
$C^{2} = A^{2} + B^{2} + 2 A B cos θ$
$13^{2} = 12^{2} + 5^{2} + 2 (12) (5) cos θ$
$cos θ = 0 \Rightarrow θ = \frac{π}{2}$

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The magnitude of vectors →A,→B and →C are respectively 12,5 and 13 units and →A+→B=→C, then the angle between →A and →B is

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The correct option is C $π / 2$
Given, $\to C = \to A + \to B$
$\to C . \to C = (\to A + \to B) . (\to A + \to B)$
$C^{2} = A^{2} + B^{2} + 2 A B cos θ$
$13^{2} = 12^{2} + 5^{2} + 2 (12) (5) cos θ$
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