# Find all values of $a$ such that $w = ai- \frac{a}{3}j$ is a unit vector

Find all values of $a$ such that $w = ai- \frac{a}{3}j$ is a unit vector.

I just need help understanding precisely what this question is asking. This is the first chapter and section of multivariable calculus.

I am going to assume then that $w$ is of length $1$, and therefore $\sqrt{(ai)^2 + (-\frac{a}{3}j)^2} = 1$

The problem with this is that I have a scalar $a$ multiplied to the unit vector $i$, and that whole thing is being squared. I have not been introduced to vector multiplication so I am going to have to assume I am not going about this the desired way (I am not simply solving for some variable $a$ using this equation, and if I am then I am unsure how to do so).

Please give me suggestions or clarifications. Am I to assume $i$ and $j$ are perpendicular unit vectors hopefully with a component of $1$ and the others are $0$? Should I consider $i$ and $j$ to take the value of $1$?

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I will assume that your book assumes that $i,j$ are the standard basis for the real plane (which I would say is a very safe assumption). $i$ in this case means exactly the vector $(1,0)$. $j$ means $(0,1)$. (This is the answer to your last question, but your middle work was a bit off).

I'd like to add that there is no one singular concept called 'vector multiplication' - though two related ideas are dot products and cross products, neither of which are necessary here.

The only think you need to know is how to find the length of a vector; all you need to do that, fortunately, is the Pythagorean Theorem.

So you should think of a right triangle with one side length $a$ and the other side length $\frac{a}{3}$, and you want to find $a$ so that the hypotenuse is of length $1$.

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So that would lead me to $a = +\sqrt{\frac{9}{10}}$ and $a = -\sqrt{\frac{9}{10}}$. Does that sound correct? –  Leonardo Apr 6 '13 at 12:12
@Leonardo, yes, that sounds right to me. Good job –  mixedmath Apr 6 '13 at 12:14
Thanks a lot, that problem has been bothering me but no longer! –  Leonardo Apr 6 '13 at 12:19
$|| w||=1$ => $\sqrt{(a)^2 +(\frac{a}{3})^2} = 1$=> $\frac{\sqrt{10}}{3}a=1$=>$a=\frac{3\sqrt{10}}{10}or a=-\frac{3\sqrt{10}}{10}$