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

The correct option is D π3 a + b + c #10;= 0 #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160 ...

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Byju's Answer

Standard XII

Mathematics

Angle between Two Line Segments

If a⃗ +b⃗ +...

Question

If $\to a + \to b + \to c = 0, ∣ ∣ \to a ∣ ∣ = 3, ∣ ∣ ∣ \to b ∣ ∣ ∣ = 5, ∣ ∣ \to c ∣ ∣ = 7,$ then the angle between $\to a & \to b$ is

\frac{π}{6}

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\frac{2 π}{3}

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\frac{5 π}{3}

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\frac{π}{3}

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Solution

The correct option is D $\frac{π}{3}$
$a + b + c = 0$ $| a | = 3, | b | = 5, | c | = 7$

$\Rightarrow \to a + \to b = - \to c$

$\Rightarrow (a + b) (a + b) = (- \to c) (- \to c)$

$\Rightarrow a^{2} + \to a . \to b + \to b . \to a + b^{- 2} = c^{2}$

$\Rightarrow 2 a . b cos θ = c^{2} - a^{2} - b^{2}$

$cos θ = \frac{c^{2} - a^{2} - b^{2}}{2 a b}$

$= \frac{49 - 9 - 25}{2 \times 3 \times 5}$

$= \frac{15}{6 \times 5}$

$= \frac{1}{2}$

$∴ θ = \frac{π}{3}$

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If →a+→b+→c=0,∣∣→a∣∣=3,∣∣∣→b∣∣∣=5,∣∣→c∣∣=7, then the angle between →a&→b is

The correct option is D π3a+b+c=0 |a|=3,|b|=5,|c|=7 ⇒→a+→b=−→c ⇒(a+b)(a+b)=(−→c)(−→c) ⇒a2+→a.→b+→b.→a+b−2=c2 ⇒2a.bcosθ=c2−a2−b2 cosθ=c2−a2−b22ab =49−9−252×3×5 =156×5 =12 ∴θ=π3

If $\to a + \to b + \to c = 0, ∣ ∣ \to a ∣ ∣ = 3, ∣ ∣ ∣ \to b ∣ ∣ ∣ = 5, ∣ ∣ \to c ∣ ∣ = 7,$ then the angle between $\to a & \to b$ is