If a and b are two vectors such that -a-b- then the greatest value of -a-b-a-b- is equal to

Let angle between a and b is θ. then, |a+b|+|a b| =√(|a|^2+|b|^2+2|a||b|cosθ)+√(|a|^2+|b|^2 2|a||b|cosθ) =√(2)|a|(√(1+cosθ)+√(1 cosθ)) [∵ |a|=|b|] =√(2)|a| ...

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Byju's Answer

Standard IX

Mathematics

Triangle Inequality

If a and b ar...

Question

If $\to a$ and $\to b$ are two vectors such that $| \to a | = | \to b |,$ then the greatest value of $| \to a + \to b | + | \to a - \to b |$ is equal to

\sqrt{2} | \to a |

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2 \sqrt{2} | \to a |

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4 | \to a |

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4 \sqrt{2} | \to a |

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Solution

The correct option is B $2 \sqrt{2} | \to a |$
Let angle between $\to a$ and $\to b$ is $θ .$
then,
$| \to a + \to b | + | \to a - \to b | = \sqrt{| \to a |^{2} + | \to b |^{2} + 2 | \to a | | \to b | cos θ} + \sqrt{| \to a |^{2} + | \to b |^{2} - 2 | \to a | | \to b | cos θ} = \sqrt{2} | \to a | (\sqrt{1 + cos θ} + \sqrt{1 - cos θ}) [∵ | \to a | = | \to b |] = \sqrt{2} | \to a | (\sqrt{2} cos \frac{θ}{2} + \sqrt{2} sin \frac{θ}{2}) = 2 | \to a | (cos \frac{θ}{2} + sin \frac{θ}{2})$
As, $cos \frac{θ}{2} + sin \frac{θ}{2} \leq \sqrt{2}$
So, $(| \to a + \to b | + | \to a - \to b |)_{m a x} = 2 \sqrt{2} | \to a |$

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If →a and →b are two vectors such that |→a|=|→b|, then the greatest value of |→a+→b|+|→a−→b| is equal to

If $\to a$ and $\to b$ are two vectors such that $| \to a | = | \to b |,$ then the greatest value of $| \to a + \to b | + | \to a - \to b |$ is equal to