Vectors a- b- and c- are such that a- - b- - c- - 0- and -a- - 3- -b- - 5 and -c- - 7- Find the angle between a- and b-

Given a⃗+b⃗+c⃗=0⃗ |a⃗|=3, |b⃗|=5, |c⃗|=7 a⃗+b⃗= c⃗ ( a⃗+b⃗ ). ( a⃗+b⃗ )+ ( c⃗ ). ( c⃗ ) a⃗.a⃗+a⃗.b⃗+a⃗.b⃗+b⃗.b⃗=|c⃗|^2 a⃗|^2+2a ...

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Byju's Answer

Standard XII

Mathematics

Applications of Dot Product

Vectors a⃗,...

Question

Vectors $\to a, \to b$ and $\to c$ are such that $\to a + \to b + \to c = \to 0$ and $| \to a | = 3, | \to b = 5$ and $| \to c | = 7$ . Find the angle between $\to a$ and $\to b .$

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Solution

Given $\to a + \to b + \to c = \to 0$
$| \to a | = 3, | \to b | = 5, & | \to c | = 7$
$\to a + \to b = - \to c$
$(\to a + \to b) . (\to a + \to b) + (- \to c) . (- \to c)$
$\to a . \to a + \to a . \to b + \to a . \to b + \to b . \to b = | \to c |^{2}$
$\to a |^{2} + 2 \to a | \to b | c o s θ + \to b |^{2} = \to c |^{2}$
$9 + 2 (3) . (5) c o s θ + 25 = 49$
$30 c o s θ = 49 - 34$
$c o s θ = \frac{15}{30}$
$c o s θ = \frac{1}{2}$
$θ = c o s^{- 1} \frac{1}{2}$
$θ = \frac{π}{3}$

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Similar questions

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Vectors →a,→b and →c are such that →a+→b+→c=→0 and |→a|=3,|→b=5 and |→c|=7. Find the angle between →a and →b.

Given →a+→b+→c=→0|→a|=3,|→b|=5,&|→c|=7→a+→b=−→c(→a+→b).(→a+→b)+(−→c).(−→c)→a.→a+→a.→b+→a.→b+→b.→b=|→c|2→a|2+2→a|→b|cosθ+→b|2=→c|29+2(3).(5)cosθ+25=4930cosθ=49−34cosθ=1530cosθ=12θ=cos−112θ=π3

Vectors $\to a, \to b$ and $\to c$ are such that $\to a + \to b + \to c = \to 0$ and $| \to a | = 3, | \to b = 5$ and $| \to c | = 7$ . Find the angle between $\to a$ and $\to b .$