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The position coordinates of a particle moving in a 3-D coordinate system is given by

$x = a\cos \omega t$

$y = a\sin \omega t$

and $z = a\omega t$

The speed of the particle is:

Option 1)

$2a\omega$
Option 2)

$\sqrt{3}a\omega$
Option 3)

$\sqrt{2}a\omega$
Option 4)

$a\omega$

Answers (1)

best_answer

Magnitude of position of vectors -

Magnitude: Its magnitude is the distance between the given point and its direction from the origin to that point.

$\vec{r}= x\hat{i}+y\hat{j}+z\hat{k}$

Magnitude : $r= \sqrt{x^{2}+y^{2}+z^{2}}$

- wherein

If $\vec{A}= 3\hat{i}-4\hat{j}+2\hat{k}$

Find its magnitude

$r= \sqrt{x^{2}+y^{2}+z^{2}}$

$= \sqrt{(3)^{2}+(-4)^{2}+(2)^{2}}$

$= \sqrt{9+16+4}$

$a= \sqrt{29}$

$V=\sqrt{V_{x}^{2}+V_{y}^{2}+V_{z}^{2}}$

$V_{x}=-a\omega \sin \omega t$

$V_{y}=a\omega \cos \omega t$

$V_{z}=a\omega$

$V=\sqrt{a^{2}\omega ^{2}+a^{2}\omega ^{2}}$

$V=\sqrt2 a\omega$

Option 1)

$2a\omega$

Option 2)

$\sqrt{3}a\omega$

Option 3)

$\sqrt{2}a\omega$
Option 4)

$a\omega$

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