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If a vector is a combination of two unit vectors, is this vector still a unit vector?

For example $\vec v_1$ and $\vec v_2$ are both unit vectors, that is, their lengths are both $1$.

Now if there is a vector $\vec f$.

$\vec f=\alpha \vec v_1+(1-\alpha )\vec v_2$ , $ 0<\alpha<1$

Is $\vec f$ also a unit vector?

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    $\begingroup$ You've accepted a good answer. I think you could have discovered it yourself if you'd drawn pictures of a few examples in the plane. That's usually a good strategy before you start trying to work with algebra and precise definitions. $\endgroup$ Oct 15 '19 at 12:43
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No. For instance, take:

$$(1,0),(0,1),\alpha=1/2$$

What is $\vec{f}$?

Note that $||\vec{f}||\neq 1$.

Is it clear? Good studies!

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We can check the conditions. So let $\vec{v}$ be a vector, being a lineair combination of 2 unit vectors $\vec{u}_1$ and $\vec{u}_2$. That is:

$\vec{v} = \alpha \vec{u}_1 + \beta \vec{u}_2$

Of course, here we suppose we are working in a 2D vectorspace spanned by the vectors $\vec{u}_1$ and $\vec{u}_2$. Taking the norm gives:

$||\vec{v}|| = \sqrt{\vec{v} \cdot \vec{v}} = \sqrt{\alpha^2 + \beta^2}$

The last gives the condition for $\vec{v}$ being a unit vector.

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