Find the components of a vector?

Question

Can I find the components of a vector while I know it's norm ? I also know it is collinear to another vector of which I know the components.

Here is an image :

Answer 1

We will denote the vector $\overrightarrow{AB}$ with $v$ and the vector $\overrightarrow{BC}$ with $w$. Since we know that both $v$ and $w$ are collinear, we can express $v$ as $v = kw$ for some constant $k$. We will spend our time now computing what the constant $k$ should be.

Recall that for any vector $x$, the dot product $x \cdot x = ||x||^2$

So, we see that $v \cdot v = ||v||^2 = ||\overrightarrow{AB}||^2 = 1.6^2 = 2.56$.

However, we may also see that since $v = kw$, we can obtain another equation:

$v \cdot v = kw \cdot kw = k^2(w\cdot w) = k^2||w||^2$

So, we have $2.56 = v \cdot v = k^2||w||^2$. However, we are given the components of the vector $w$. In fact, we know that $w = (40\cos(60), 40\sin(60))$ (and I will assume these are in degrees), we find that $||w||^2 = (40\cos(60))^2 + (40\sin(60))^2 = 1600$. Thus, we have $$2.56 = k^2 (1600) \implies k^2 = \frac{1}{625} \implies k = \pm \frac{1}{25}$$ In order to figure out which sign $k$ will be, we notice that the notation $\overrightarrow{AB}$ and $\overrightarrow{BC}$ along with the picture you have provided are highly suggestive of the fact that both vectors will "point" in the same direction. From this information, we can conclude that $k = + \frac{1}{25}$. If we chose the negative sign, the vector $v$ would point in the opposite direction as $w$. So, we have $v = \frac{1}{25}w = \frac{1}{25}(40\cos(60), 40\sin(60)) = (1.6\cos(60), 1.6\sin(60))$