Two vectors A-B- are such that - A - B - Magnitude of A - B is -3 times of magnitude of A - B - The angle between A - B is A- 120-B- 60-C- 30-D- 90-

The correct option is B 60^∘Let angle between A⃗& B⃗ is θ |A⃗ + B⃗| = √(A^2 + B^2 + 2AB cos θ) |A⃗ B⃗| = √(A^2 + B^2 2AB cos θ) √(A^2 + B^2 + 2AB cos θ) ...

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Standard XII

Mathematics

Substitution Method to Remove Indeterminate Form

Two vectors A...

Question

Two vectors $\to A & \to B$ are such that $| \to A | = | \to B |$ . Magnitude of ( $\to A + \to B$ ) is $\sqrt{3}$ times of magnitude of ( $\to A - \to B$ ). The angle between $\to A & \to B$ is

120^{\circ}

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60^{\circ}

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30^{\circ}

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90^{\circ}

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Solution

The correct option is B $60^{\circ}$
Let angle between $\to A & \to B$ is $θ$
$| \to A + \to B | = \sqrt{A^{2} + B^{2} + 2 A B c o s θ}$
$| \to A - \to B | = \sqrt{A^{2} + B^{2} - 2 A B c o s θ}$
$\sqrt{A^{2} + B^{2} + 2 A B c o s θ} = \sqrt{3} \sqrt{A^{2} + B^{2} - 2 A B c o s θ}$
$A^{2} + B^{2} + 2 A B c o s θ = 3 A^{2} + 3 B^{2} - 6 A B c o s θ$
$8 A B c o s θ = 2 A^{2} + 2 B^{2}$
$4 c o s θ = \frac{A^{2} + B^{2}}{A B}$
$4 c o s θ = 2$
$c o s θ = \frac{2}{4}$
$θ = 60^{\circ}$
Hence, option $B$ is the correct answer.

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Two vectors →A & →B are such that |→A|=|→B|. Magnitude of (→A+→B) is √3 times of magnitude of ( →A−→B). The angle between →A & →B is

Two vectors $\to A & \to B$ are such that $| \to A | = | \to B |$ . Magnitude of ( $\to A + \to B$ ) is $\sqrt{3}$ times of magnitude of ( $\to A - \to B$ ). The angle between $\to A & \to B$ is