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A vector $\vec{B}$ has a magnitude $B$ and and a unit vector $\hat{B}$ in the direction of $B$ then which of the following are correct

1) $\vec{B} .\hat{B} = B$

2) $\hat{B} = \frac {\vec{B}} {B}$

3) $\vec{B}.\vec{B} = B^2$

4) $B = \frac {\vec{B}} {\hat{B}}$

So I was attempting this question and I got my answers as 1), 2), 3), 4)

Now according to my book the answer is 1), 2), 3).

I don't understand why. If $\hat{B} = \frac {\vec{B}} {B}$ is correct then why not $B = \frac {\vec{B}} {\hat{B}}$ because in this we have just interchanged the denominators through cross multiplication right.

Please correct if I am wrong and please justify your answer.

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1 Answer 1

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You can't divide vectors. It's just not defined.

As a bit of clarification: you could just as easily say that $\hat{B} = \frac{1}{B} I \vec{B}$, and from that conclude that $\frac{\hat{B}}{\vec{B}} = \frac{1}{B} I$, where I is the unit matrix. Multiplication just doesn't work nicely enough here to allow unambigious division.

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  • $\begingroup$ Oh thanks...... $\endgroup$
    – user663795
    May 14, 2019 at 18:09
  • $\begingroup$ As a bit of clarification: you could just as easily say that $\hat{B} = \frac{1}{B} I \vec{B}$, and from that conclude that $\frac{ \hat{B}}{\vec{B}} = \frac{1}{B} I$, where $I$ is the unit matrix. Multiplication just doesn't work nicely enough here to allow unambigious division. $\endgroup$ May 14, 2019 at 18:12
  • $\begingroup$ @AlexanderGeldhof please include this in the answer itself. $\endgroup$ May 14, 2019 at 18:31
  • $\begingroup$ Fair enough, @MohammadZuhairKhan. $\endgroup$ May 15, 2019 at 12:28

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