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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)

best_answer

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

Posted by

prateek

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