Given that C - A - B- Also- the magnitude of A- B and C are 12- 5 and 13 units respectively- The angle between A and B is

If C = A + B, then magnitude of C and A + B are same Hence, |C|=|A + B| |C| = √(A^2 + B^2 + 2AB cosθ) 13 = √(12^2 + 5^2 + 2 × 12 × 5 cosθ) ⇒ 169 = 144 + 25 + 2 ...

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Standard XIII

Physics

Vector Component

Given that C ...

Question

Given that $\to C = \to A + \to B$ . Also, the magnitude of $\to A, \to B$ and $\to C$ are $12, 5$ and $13$ units respectively. The angle between $\to A$ and $\to B$ is

0^{\circ}

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\frac{π}{4}

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\frac{π}{2}

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π

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Solution

The correct option is C $\frac{π}{2}$
If $\to C = \to A + \to B$ , then magnitude of $\to C$ and $\to A + \to B$ are same
Hence, $| \to C | = | \to A + \to B |$
$| \to C | = \sqrt{A^{2} + B^{2} + 2 A B cos θ}$
$13 = \sqrt{12^{2} + 5^{2} + 2 \times 12 \times 5 cos θ}$
$\Rightarrow 169 = 144 + 25 + 2 \times 12 \times 5 cos θ$
$cos θ = 0 \Rightarrow θ = \frac{π}{2}$

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