The maximum value of -A-B- is-

The correct option is B |A⃗| + |B⃗| |A B| = √(A^2+B^2 2A.Bcosθ) Maximum value occurs when θ = π #160;cos θ = 1And #160; |A B| = ...

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Standard XII

Mathematics

Trigonometric Ratios Using Right Angled Triangle

The maximum v...

Question

The maximum value of $| (\to A - \to B) |$ is:

| \to A | - | \to B |

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| \to A | + | \to B |

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A^{2} + B^{2}

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A^{2} - B^{2}

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Solution

The correct option is B $| \to A | + | \to B |$
$| \to A - \to B | = \sqrt{A^{2} + B^{2} - 2 A . B c o s θ}$
Maximum value occurs when $θ = π$
cos $θ$ = -1
And $| \to A - \to B | = \sqrt{A^{2} + B^{2} + 2 A . B . c o s θ} = \sqrt{(| A | + | B |)^{2}} = | A | + | B |$

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The maximum value of |(→A−→B)| is:

The correct option is B |→A|+|→B||→A−→B|=√A2+B2−2A.BcosθMaximum value occurs when θ=π cos θ = -1And |→A−→B|=√A2+B2+2A.B.cosθ=√(|A|+|B|)2=|A|+|B|

The maximum value of $| (\to A - \to B) |$ is:

The correct option is B $| \to A | + | \to B |$
$| \to A - \to B | = \sqrt{A^{2} + B^{2} - 2 A . B c o s θ}$
Maximum value occurs when $θ = π$
cos $θ$ = -1
And $| \to A - \to B | = \sqrt{A^{2} + B^{2} + 2 A . B . c o s θ} = \sqrt{(| A | + | B |)^{2}} = | A | + | B |$