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JEE Main 202120 JulyMorning Shift
Exam
JEE Main
Year
2021
Date
20 July
Shift
Morning Shift
Subject
Physics
Chapter
Mathematics in Physics

JEE Main 2021 · 20 July · Morning Shift · Physics · Mathematics in Physics

If → A and → B are two vectors satisfying the relation → A · → B =...

If A→\overset{\to}{A} and B→\overset{\to}{B} are two vectors satisfying the relation A→⋅B→=∣A→×B→∣\overset{\to}{A} \cdot \overset{\to}{B} = | \overset{\to}{A} \times \overset{\to}{B} |. Then the value of ∣A→−B→∣| \overset{\to}{A} - \overset{\to}{B} | will be:

Options

  1. AA2+B2\sqrt{{A}^{2} + {B}^{2}}
  2. BA2+B2+2AB\sqrt{{A}^{2} + {B}^{2} + \sqrt{2}AB}
  3. CA2+B2+2AB\sqrt{{A}^{2} + {B}^{2} + 2AB}
  4. DA2+B2−2AB\sqrt{{A}^{2} + {B}^{2} - \sqrt{2}AB}

Correct answer

D. A2+B2−2AB\sqrt{{A}^{2} + {B}^{2} - \sqrt{2}AB}

Step-by-step solution

A→⋅B→=∣A→×B→∣\overset{\to}{A} \cdot \overset{\to}{B} = | \overset{\to}{A} \times \overset{\to}{B} |
ABcosθ=ABsinθ⇒θ=45°ABcos \theta = ABsin \theta ⇒ \theta = 45 °
∣A→−B→∣=A2+B2−2ABcos45°| \overset{\to}{A} - \overset{\to}{B} | = \sqrt{{A}^{2} + {B}^{2} - 2ABcos45 °}
=A2+B2−2AB= \sqrt{{A}^{2} + {B}^{2} - \sqrt{2}AB}
Hence option (4).( 4 ) .
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