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A vector $\vec{A}$ is rotated by a small angle $\Delta \Theta$ radians $\left ( \Delta \Theta < < 1 \right )$ to get a new vector $\vec{B}\left$ .In that case $\left | \vec{B}-\vec{A} \right |$ is :

Option 1)
$0$

Option 2)
$\left | \vec{A} \right |\left ( 1-\frac{\Delta \Theta ^{2}}{2} \right )$

Option 3)
$\left | \vec{A} \right |\Delta \Theta$

Option 4)
$\left | \vec{B} \right |\Delta \Theta\left -| \vec{A} \right |$

Answers (1)

best_answer

As we discussed in

Triangle law of vector Addition -

If two vector are represented by both magnitude and direction by two sides of triangle taken in same order then their resultant is represented by 3^rd side of triangle.

- wherein

Represents triangle law of vector Addition

By Triangle Rule

$\vec{A}+\vec{C}= \vec{B}$

$\vec{B}-\vec{A}= \vec{C}$

$\left |\vec{B}-\vec{A} \right |=\left | \vec{C} \right |$ $=\left | \vec{B} \right |\sin \Theta (\because \Delta \Theta < < 1)$

$\left | \vec{B}-\vec{A} \right |=\left | \vec{B}\right | \Delta\Theta\ \; (\because sin\Delta \Theta =\Delta \Theta)$

Again $\left | \vec{B}\right |cos\Delta \Theta =\left | \vec{A}\right |$

$\left | \vec{B}\right | =\left | \vec{A}\right |$ $(\because cos\Delta \Theta \simeq 1)$

So $\left | \vec{B-A}\right | =\left | \vec{B}\right |\Delta \Theta =\left | \vec{A}\right |.\Delta \Theta$

Option 1)

$0$

Incorrect

Option 2)

$\left | \vec{A} \right |\left ( 1-\frac{\Delta \Theta ^{2}}{2} \right )$

Incorrect

Option 3)

$\left | \vec{A} \right |\Delta \Theta$

Correct

Option 4)

$\left | \vec{B} \right |\Delta \Theta\left -| \vec{A} \right |$

Incorrect

Posted by

prateek

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