If a-b and c are vectors such that a-b-c-0 and -a-7-b-5 and -c-3- then the angle between b and c is

A+b+c=0 b+c= a squaring on both sides 25+9+2b·c=49⇒2 | b|| c|cosθ =49 34=15⇒ 2× 5× 3cosθ =15⇒cosθ =12⇒θ =π3=60^∘ ...

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Byju's Answer

Standard XII

Mathematics

Test for Collinearity of Vectors

If a,b and c ...

Question

If $\to a$ , $\to b$ and $\to c$ are vectors such that $\to a +$ $\to b +$ $\to c$ $= \to 0$ and $| \to a | = 7,$ $| \to b | = 5$ and $| \to c | = 3,$ then the angle between $\to b$ and $\to c$ is

60^{o}

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30^{o}

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45^{o}

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90^{o}

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Solution

The correct option is A $60^{o}$
$\to a + \to b + \to c = \to 0$
$\to b + \to c = - \to a$
squaring on both sides
$25 + 9 + 2 \to b \cdot \to c = 49$ $\Rightarrow 2 ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣ cos θ = 49 - 34 = 15$ $\Rightarrow 2 \times 5 \times 3 cos θ = 15$ $\Rightarrow cos θ = \frac{1}{2} \Rightarrow θ = \frac{π}{3} = 60^{\circ}$

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Test for Collinearity of Vectors

Standard XII Mathematics

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If →a,→b and →c are vectors such that →a+→b+→c=→0 and |→a|=7,|→b|=5 and |→c|=3, then the angle between →b and →c is

The correct option is A 60o→a+→b+→c=→0 →b+→c=−→a squaring on both sides 25+9+2→b⋅→c=49⇒2∣∣∣→b∣∣∣∣∣→c∣∣cosθ=49−34=15⇒2×5×3cosθ=15⇒cosθ=12⇒θ=π3=60∘

If $\to a$ , $\to b$ and $\to c$ are vectors such that $\to a +$ $\to b +$ $\to c$ $= \to 0$ and $| \to a | = 7,$ $| \to b | = 5$ and $| \to c | = 3,$ then the angle between $\to b$ and $\to c$ is