If -a-b-a-b-1- then -a-b- is equal to

The correct option is D √(3) |a+b|=1 ⇒ a^2+b^2+2ab=1 [squaring both sides] ⇒ 1+1+2(1)(1)cosθ=1(θ is angle between a and b) cosθ= 12→(1) ...

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Standard XII

Mathematics

Direction Cosines

If |a|=|b|=...

Question

If $| \to a | = | \to b | = | \to a + \to b |$ =1, then $| \to a - \to b |$ is equal to

1

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\sqrt{2}

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\sqrt{3}

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None of these

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Solution

The correct option is D $\sqrt{3}$
$| a + b | = 1 \Rightarrow a^{2} + b^{2} + 2 a b = 1$
[squaring both sides]
$\Rightarrow 1 + 1 + 2 (1) (1) cos θ = 1 (θ$ is angle between a and b)
$cos θ = \frac{- 1}{2} \to (1)$
Now value of $(a - b)$
Squaring $(a - b)$ we get
$a^{2} + b^{2} - 2 a b cos θ 1 + 1 - 2 \times 1 \times 1 \times (\frac{- 1}{2}) = 3$
Now take root to fetch $| a - b | = \sqrt{3}$

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The correct option is D √3|a+b|=1⇒a2+b2+2ab=1[squaring both sides]⇒1+1+2(1)(1)cosθ=1(θ is angle between a and b)cosθ=−12→(1)Now value of (a−b)Squaring (a−b) we geta2+b2−2abcosθ1+1−2×1×1×(−12)=3Now take root to fetch |a−b|=√3

If $| \to a | = | \to b | = | \to a + \to b |$ =1, then $| \to a - \to b |$ is equal to