If - F 1- F 2- F 1- F 2- then - F 1- F 2- isA- - F 1- F 2-B- - F 1-2- F 2-2C- - F 1-2- F 2-2- F 1- F 2-2D- - F 1-2- F 2-2-2- F 1- F 2-

The correct option is D √( | F_1 |^2+ | F_2 |^2+√(2) | F_1 | | F_2 |)Given, | F_1×F_2 |=F_1.F_2 We know that, ⇒ |F_1×F_2 |= | F_1 | | F_2 | sinθ ⇒F_1. F_2= | ...

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Byju's Answer

Standard XII

Mathematics

Applications of Dot Product

If | F 1× F 2...

Question

If $∣ ∣ ∣_{1} \times_{2} ∣ ∣ ∣ =_{1} ._{2}$ , then $∣ ∣ ∣_{1} +_{2} ∣ ∣ ∣$ is

∣ ∣ ∣_{1} ∣ ∣ ∣ + ∣ ∣ ∣_{2} ∣ ∣ ∣

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\sqrt{{∣ ∣ ∣_{1} ∣ ∣ ∣}^{2} + {∣ ∣ ∣_{2} ∣ ∣ ∣}^{2}}

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\sqrt{{∣ ∣ ∣_{1} ∣ ∣ ∣}^{2} + {∣ ∣ ∣_{2} ∣ ∣ ∣}^{2} + \frac{∣ ∣ ∣_{1} ∣ ∣ ∣ ∣ ∣ ∣_{2} ∣ ∣ ∣}{\sqrt{2}}}

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\sqrt{{∣ ∣ ∣_{1} ∣ ∣ ∣}^{2} + {∣ ∣ ∣_{2} ∣ ∣ ∣}^{2} + \sqrt{2} ∣ ∣ ∣_{1} ∣ ∣ ∣ ∣ ∣ ∣_{2} ∣ ∣ ∣}

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Solution

The correct option is D $\sqrt{{∣ ∣ ∣_{1} ∣ ∣ ∣}^{2} + {∣ ∣ ∣_{2} ∣ ∣ ∣}^{2} + \sqrt{2} ∣ ∣ ∣_{1} ∣ ∣ ∣ ∣ ∣ ∣_{2} ∣ ∣ ∣}$
Given, $∣ ∣ ∣_{1} \times_{2} ∣ ∣ ∣ =_{1} ._{2}$

We know that, $\Rightarrow ∣ ∣ ∣_{1} \times_{2} ∣ ∣ ∣ = ∣ ∣ ∣_{1} ∣ ∣ ∣ ∣ ∣ ∣_{2} ∣ ∣ ∣ sin θ$
$\Rightarrow_{1} ._{2} = ∣ ∣ ∣_{1} ∣ ∣ ∣ ∣ ∣ ∣_{2} ∣ ∣ ∣ cos θ$

$∴ ∣ ∣ ∣_{1} ∣ ∣ ∣ ∣ ∣ ∣_{2} ∣ ∣ ∣ sin θ = ∣ ∣ ∣_{1} ∣ ∣ ∣ ∣ ∣ ∣_{2} ∣ ∣ ∣ cos θ$
$\Rightarrow tan θ = 1 \Rightarrow θ = 45^{\circ}$

$∴ ∣ ∣ ∣_{1} +_{2} ∣ ∣ ∣ = \sqrt{{∣ ∣ ∣_{1} ∣ ∣ ∣}^{2} + {∣ ∣ ∣_{2} ∣ ∣ ∣}^{2} + 2 ∣ ∣ ∣_{1} ∣ ∣ ∣ ∣ ∣ ∣_{2} ∣ ∣ ∣ cos 45^{\circ}}$
$= \sqrt{{∣ ∣ ∣_{1} ∣ ∣ ∣}^{2} + {∣ ∣ ∣_{2} ∣ ∣ ∣}^{2} + \sqrt{2} ∣ ∣ ∣_{1} ∣ ∣ ∣ ∣ ∣ ∣_{2} ∣ ∣ ∣}$

Hence option D is the correct answer

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