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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Mathematics in Physics

JEE Main · Physics · Mathematics in Physics

Force F applied on a body is written as F =( n F ) n + G , where n...

Force F⃗\vec{F} applied on a body is written as F→=(n^F^)n^+G→\overrightarrow{\mathrm{F}}=(\hat{\mathrm{n}} \hat{\mathrm{F}}) \hat{\mathrm{n}}+\overrightarrow{\mathrm{G}}, where n^\hat{\mathrm{n}} is a unit vector The vector G⃗\vec{G} is equal to

Options

  1. An^×F→\hat{\mathrm{n}} \times \overrightarrow{\mathrm{F}}
  2. Bn^×(n^×F→)\hat{\mathrm{n}} \times(\hat{\mathrm{n}} \times \overrightarrow{\mathrm{F}})
  3. C(n^×F→)×F→/∣F→∣(\hat{\mathrm{n}} \times \overrightarrow{\mathrm{F}}) \times \overrightarrow{\mathrm{F}} /|\overrightarrow{\mathrm{F}}|
  4. D(n^×F→)×n^(\hat{\mathrm{n}} \times \overrightarrow{\mathrm{F}}) \times \hat{\mathrm{n}}

Correct answer

D. (n^×F→)×n^(\hat{\mathrm{n}} \times \overrightarrow{\mathrm{F}}) \times \hat{\mathrm{n}}

Step-by-step solution

(n^×F→)×n^=−[n^×(n^×F→)]=−[n^(n^⋅F→)−F→(n^⋅n^)]=F→−n^(n^⋅F→)]=G→(\hat{\mathrm{n}} \times \overrightarrow{\mathrm{F}}) \times \hat{\mathrm{n}}=-[\hat{\mathrm{n}} \times(\hat{\mathrm{n}} \times \overrightarrow{\mathrm{F}})]=-[\hat{\mathrm{n}}(\hat{\mathrm{n}} \cdot \overrightarrow{\mathrm{F}})-\overrightarrow{\mathrm{F}}(\hat{\mathrm{n}} \cdot \hat{\mathrm{n}})]=\overrightarrow{\mathrm{F}}-\hat{\mathrm{n}}(\hat{\mathrm{n}} \cdot \overrightarrow{\mathrm{F}})]=\overrightarrow{\mathrm{G}}
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