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I stumbled upon a question which states the following -

If vector $A = 0.6\bf\hat{i} + N\bf\hat{j}$ is a unit vector, find the value of $N$.

On solving, the value for $N$ would be 0.8 . But my real question is - Is this possible?

How adding two vectors which have magnitudes (number × unit), gave a resultant vector having magnitude 1 (without any unit)?

Am I missing something here ?

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    $\begingroup$ There are no units specified in the problem. Anyway the magnitude of a unit vector is $1\times$unit. $\endgroup$
    – mike stone
    Commented Jul 1 at 17:56
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    $\begingroup$ just sketch a unit vector on paper, I am sure you find many combinations of two vectors which get it. $\endgroup$
    – trula
    Commented Jul 1 at 17:57
  • $\begingroup$ In fact, an infinite number of combinations. $\endgroup$
    – Jon Custer
    Commented Jul 1 at 18:03
  • $\begingroup$ @mikestone Can you explain what you meant by 1 unit ? As of my understanding Its magnitude is just 1 without any units. $\endgroup$
    – Agog 0615
    Commented Jul 1 at 18:04
  • $\begingroup$ A proper answer with examples would be of great help . Thank you. $\endgroup$
    – Agog 0615
    Commented Jul 1 at 18:06

2 Answers 2

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Your vector is $\mathbf{A} = (0.6, N)$, and you want to find $N$ such that $\| {A} \| = 1$. The equations for that is $0.6^2 + N^2 = 1$. You can tell at this level that $N^2$ and therefore $N$ must be unitless because of the addition to the also dimensionless $0.36$. It had to be this way since $\mathbf{A}$ itself was specified to be a unit vector (and therefore dimensionless).

As regards your specific question:

How adding two vectors which have magnitudes (number × unit), gave a resultant vector having magnitude 1(without any unit) ?

Am I missing something here ?

The answer is that you don't. What you missed is that you didn't have anything in your problem of the form "number x unit" from the start.

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  • $\begingroup$ Okay , now I get it. But again , is this correct? I mean I didn't get the idea how two vectors here are unitless ? Does it mean they are already a fraction of base vectors ( 0.6 of i^ and 0.8 of j^ ) ? Am I right in thinking this ? $\endgroup$
    – Agog 0615
    Commented Jul 1 at 18:23
  • $\begingroup$ The vectors ${\bf i}$ and ${\bf j}$ are unit vectors. by definition $\endgroup$
    – mike stone
    Commented Jul 1 at 18:24
  • $\begingroup$ Is what correct? The unit vectors are $\mathbf{\hat{i}}$ and $\mathbf{\hat{j}}$. Those have not units. For the definition of $\mathbf{A}$, the coefficients also have no units for the reasons in my answer. I don't think it's any more or less than that. $\endgroup$
    – Brick
    Commented Jul 1 at 18:25
  • $\begingroup$ @Brick didn't get the complete gist of your answer previously. But now I understand . Thank you so much for your answer. $\endgroup$
    – Agog 0615
    Commented Jul 1 at 18:32
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A general 2d vector will have the form $$ \mathbf{v} = a \mathbf{i} + b \mathbf{j} $$ where $\mathbf{i} \cdot \mathbf{i} = \mathbf{j} \cdot \mathbf{j} = 1$ and $\mathbf{i} \cdot \mathbf{j} = 0$.

The length of this vector is $$ \ell = \sqrt{\mathbf{v}\cdot\mathbf{v}} = \sqrt{a^2+b^2} $$ From here we can say a few things.

  • If $a=1$ and $b\neq 0$ (or vice versa), then $\ell >1$, so $\mathbf{v}$ will not be a unit vector.
  • If $a$ and $b$ have units, they should be the same unit -- call it $U$. Then $\ell$ will also have units of $U$.
  • $\mathbf{v}$ will be a unit vector if $a^2+b^2=1$. One common way to enforce this constraint is to introduce a parameter $\theta$, and then do a change of variables $a=\sin \theta$ and $b=\cos\theta$, so $\mathbf{v}=\sin\theta\mathbf{i} + \cos\theta\mathbf{j}$.
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