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2 years ago

If A=3i +2j+3k ,find the magnitude of A+B and A-B

Physics

1 answer:

omeli [17]2 years ago

4 0

<u>Since vector B was not specified, I'll assume one at random. You can later answer your own question.</u>

Answer:

$\mid\mid \vec A+\vec B \mid \mid=\sqrt{105}$

$\mid\mid \vec A-\vec B \mid \mid=\sqrt{149}$

Explanation:

Given:

$\vec A=3\hat i +2\hat j+3\hat k$

And (assumed):

$\vec B=-5\hat i +8\hat j-4\hat k$

Find the magnitude of

$\vec A+\vec B$

$\vec A-\vec B$

Given a vector

$\vec P=x\hat i +y\hat j+z\hat k$

The magnitude of the vector is:

$\mid\mid \vec P\mid \mid=\sqrt{x^2+y^2+z^2}$

First part:

$\vec A+\vec B =3\hat i +2\hat j+3\hat k-5\hat i +8\hat j-4\hat k$

$\vec A+\vec B =-2\hat i +10\hat j-\hat k$

The magnitude of the sum is:

$\mid\mid \vec A+\vec B \mid \mid=\sqrt{(-2)^2+10^2+(-1)^2}=\sqrt{4+100+1}$

$\mathbf{\mid\mid \vec A+\vec B \mid \mid=\sqrt{105}}$

Second part:

$\vec A-\vec B =3\hat i +2\hat j+3\hat k-(-5\hat i +8\hat j-4\hat k)$

$\vec A-\vec B =3\hat i +2\hat j+3\hat k+5\hat i -8\hat j+4\hat k$

$\vec A-\vec B =8\hat i -6\hat j+7\hat k$

The magnitude of the difference is:

$\mid\mid \vec A-\vec B \mid \mid=\sqrt{8^2+(-6)^2+7^2}=\sqrt{64+36+49}$

$\mathbf{\mid\mid \vec A-\vec B \mid \mid=\sqrt{149}}$

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2 years ago

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Therefore on solving the above differential equation we get,

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where A(t)=A₀ $e^{\frac{-bt}{2m}}$

A₀ is the amplitude at t=0 and

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$w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

Now coming to the problem,

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applying logarithm on both sides

⇒ $\frac{bt}{2m}=ln(8)$

⇒ $t=\frac{2m*ln(8)}{b}$

substituting the values

$t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)$

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3 years ago

If A=3i +2j+3k ,find the magnitude of A+B and A-B​

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