Let a- b and c be three vectors such that a - b - c - 0- and -a- - 10- -b- - 6 and -c- - 14- What is the angle between a and b

The correct option is A 60^o a+b+c=0 ∴a+b= c ∴ (a+b).(a+b)=( c).( c) ∴ |a|^2+|b|^2+2|a||b|cosθ=|c|^2 ∴ 100+36+120cosθ=196 ∴ ...

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Byju's Answer

Standard XII

Mathematics

Applications of Dot Product

Let a, b an...

Question

Let $\to a, \to b$ and $\to c$ be three vectors such that $\to a + \to b + \to c = \to 0$ , and $| \to a | = 10, | \to b | = 6$ and $| \to c | = 14$ . What is the angle between $\to a$ and $\to b$

30^{o}

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45^{o}

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60^{o}

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75^{o}

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Solution

The correct option is A $60^{o}$
$\to a + \to b + \to c = 0$
$∴ \to a + \to b = - \to c$

$∴ (\to a + \to b) . (\to a + \to b) = (- \to c) . (- \to c)$

$∴ | \to a |^{2} + | \to b |^{2} + 2 | \to a | | \to b | c o s θ = | \to c |^{2}$

$∴ 100 + 36 + 120 c o s θ = 196$

$∴ c o s θ = \frac{1}{2}$

$∴ θ = 60^{0}$

The answer is option (C)

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Let →a,→b and →c be three vectors such that →a+→b+→c=→0, and |→a|=10,|→b|=6 and |→c|=14. What is the angle between →a and →b

The correct option is A 60o→a+→b+→c=0∴→a+→b=−→c∴(→a+→b).(→a+→b)=(−→c).(−→c)∴|→a|2+|→b|2+2|→a||→b|cosθ=|→c|2∴100+36+120cosθ=196∴cosθ=12∴θ=600The answer is option (C)

Let $\to a, \to b$ and $\to c$ be three vectors such that $\to a + \to b + \to c = \to 0$ , and $| \to a | = 10, | \to b | = 6$ and $| \to c | = 14$ . What is the angle between $\to a$ and $\to b$