If the magnitude of vectors A- B- and C- are 12- 5 and 13 units- respectively- and A- B-C- the angle between vectors A- and B- is-

The correct option is B π/2 A⃗ +B⃗ = C⃗ Thus, |A⃗ +B⃗ | =|C⃗ |= 13 or, √( A ^ 2 + B ^ 2 +2AB.cosθ #160;) =13 which gives, 144+ 25+ 2a ...

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Byju's Answer

Standard XII

Physics

Addition and Subtraction in Unit Vector Notation

If the magnit...

Question

If the magnitude of vectors $\to A, \to B$ and $\to C$ are $12, 5$ and $13$ units, respectively, and $\to A + \to B = \to C$ , the angle between vectors $\to A a n d \to B$ is:

\frac{π}{4}

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\frac{π}{2}

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π

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Solution

The correct option is B $\frac{π}{2}$
$\to A + \to B = \to C$
Thus, $| \to A + \to B | = | \to C | = 13$
or, $\sqrt{A^{2} + B^{2} + 2 A B . c o s θ} = 13$
which gives, $144 + 25 + 2 a b . c o s θ = 169 o r c o s θ = 0.$
Thus, $θ = π / 2$

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