If vector \vec{a}+\vec{b} is perpendicular to  \vec{b} and \vec{2b}+\vec{a} is perpendicular to \vec{a}, then

  • Option 1)

    \left | \vec{a} \right |=\sqrt{2}\left | \vec{b} \right |

  • Option 2)

    \left | \vec{a} \right |={2}\left | \vec{b} \right |

  • Option 3)

    \left | \vec{b} \right |=\sqrt{2}\left | \vec{a} \right |

  • Option 4)

    \left | \vec{a} \right |=\left | \vec{b} \right |

 

Answers (1)

Use concept ID

Scalar Product of two vectors -

\vec{a}.\vec{b}> 0 \:an\: acute\: angle

\vec{a}.\vec{b}< 0 \:an\: obtuse\: angle

\vec{a}.\vec{b}= 0 \:a\:right\: angle

- wherein

\Theta  is the angle between the vectors \vec{a}\:and\:\vec{b}

 

 \left ( \vec{a} \right+\vec{b} ).\vec{b}=0 \Rightarrow \vec{a}.\vec{b}+\vec{b}.\vec{b}=0

and   \left ( 2\vec{b}+\vec{a} \right).\vec{a}=0\Rightarrow 2\vec{b}.\vec{a}+\vec{a}.\vec{a}=0

\therefore \vec{a}.\vec{b}+\left | b \right |^{2}=2\vec{a}.\vec{b}+\left | a \right |^{2}

\left | \vec{b} \right |^{2}-\left | \vec{a} \right |^{2}=\vec{a}.\vec{b}

\therefore \left | b \right |^{2}-\left | a \right |^{2}+\left | b \right |^{2}=0

\left | a \right |^{2}=2\left | b \right |^{2}

\left | \vec{a} \right |=\sqrt{2}\left | \vec{b} \right |

 

 


Option 1)

\left | \vec{a} \right |=\sqrt{2}\left | \vec{b} \right |

Correct option

Option 2)

\left | \vec{a} \right |={2}\left | \vec{b} \right |

Incorrect option

Option 3)

\left | \vec{b} \right |=\sqrt{2}\left | \vec{a} \right |

Incorrect option

Option 4)

\left | \vec{a} \right |=\left | \vec{b} \right |

Incorrect option

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