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

The correct option is C π3 a⃗ + b⃗ + c⃗ = 0 a⃗ + b⃗ = #160; c⃗ ( a⃗ + b⃗)^2 = ( c⃗)^2 ( a⃗ + b⃗)( a⃗ + b⃗) = | c⃗|^2 a⃗a⃗ ...

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Byju's Answer

Standard XII

Mathematics

Relation between Radian and Degree

If a+b+c= ...

Question

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

\frac{π}{\begin{matrix} 6 \end{matrix}}

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\frac{2 π}{3}

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\frac{π}{3}

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\frac{5 π}{3}

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Solution

The correct option is C $\frac{π}{3}$
$\to a + \to b + \to c = 0$
$\to a + \to b = - \to c$
${(\to a + \to b)}^{2} = {(- \to c)}^{2}$
$(\to a + \to b) (\to a + \to b) = {| \to c |}^{2}$
$\to a \to a + \to a \to b + \to b \to a + \to b \to b = {| \to c |}^{2}$
${(3)}^{2} + 2 \to a \to b + {(5)}^{2} = {(7)}^{2}$
$2 \to a \to b = 15$
$\to a \to b = \frac{15}{2}$
$| \to a | \times ∣ ∣ \to b ∣ ∣ cos θ = \frac{15}{2}$
$cos θ = \frac{1}{2}$
$θ = \frac{π}{3}$

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If a+b+c=0 and ∣∣→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+→b)(→a+→b)=|→c|2→a→a+→a→b+→b→a+→b→b=|→c|2(3)2+2→a→b+(5)2=(7)22→a→b=15→a→b=152|→a|×∣∣→b∣∣cosθ=152cosθ=12 θ=π3

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