Suppose a-b-c-0-a-3-b-5-c-7- then the angle between a- and b- is

The correct option is C π/3 nbsp; nbsp; nbsp;a+b+c=0 ⇒a+b= c ⇒ |a+b|^2=|c|^2 ⇒ nbsp;|a|^2+2a· nbsp;b +|b|^2=|c|^2 ⇒ 9+ 2× 3 × 5 cosθ+25=49 ⇒cosθ=1/2 ...

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Byju's Answer

Standard XII

Physics

Vector Component

Suppose a⃗+b⃗...

Question

Suppose $\to a + \to b + \to c = \to 0,$ $| \to a | = 3, | \to b | = 5, | \to c | = 7,$ then the angle between $\to a$ and $\to b$ is

π

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\frac{π}{2}

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\frac{π}{3}

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\frac{π}{4}

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Solution

The correct option is C $\frac{π}{3}$
$\to a + \to b + \to c = \to 0$
$\Rightarrow \to a + \to b = - \to c$
$\Rightarrow | \to a + \to b |^{2} = | \to c |^{2}$
$\Rightarrow | \to a |^{2} + 2 \to a \cdot \to b + | \to b |^{2} = | \to c |^{2}$
$\Rightarrow 9 + 2 \times 3 \times 5 cos θ + 25 = 49$
$\Rightarrow cos θ = \frac{1}{2}$
$\Rightarrow θ = \frac{π}{3}$

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Suppose →a+→b+→c=→0, |→a|=3,|→b|=5,|→c|=7, then the angle between →a and →b is

The correct option is C π3 →a+→b+→c=→0 ⇒→a+→b=−→c ⇒|→a+→b|2=|→c|2 ⇒|→a|2+2→a⋅→b+|→b|2=|→c|2 ⇒9+2×3×5cosθ+25=49 ⇒cosθ=12 ⇒θ=π3

Suppose $\to a + \to b + \to c = \to 0,$ $| \to a | = 3, | \to b | = 5, | \to c | = 7,$ then the angle between $\to a$ and $\to b$ is