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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 :

Here is an image

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  • $\begingroup$ You need to formulate an equation for the second vector. First, write the equation for the fact that it's colinear with another vector. Hint: It means that there is a constant $k$ so that (you fill the rest) ... $\endgroup$ – Matti P. Apr 12 at 7:12
  • $\begingroup$ So I should use the formula (a, b) = k*(c ,d) ? but that is not enough right ? $\endgroup$ – Jonathcraft Apr 12 at 7:34
  • $\begingroup$ Yes, very good, and then another equation, you need to incorporate the length of the new vector, and therefore find the coefficient $k$. And then you're done. $\endgroup$ – Matti P. Apr 12 at 7:35
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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 $v = \frac{1}{25}w = \frac{1}{25}(40\cos(60), 40\sin(60)) = (1.6\cos(60), 1.6\sin(60))$

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    $\begingroup$ Thanks a lot, I didn't think of it like that. Plus it is easy to implement into a program $\endgroup$ – Jonathcraft Apr 12 at 8:09

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