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

  • $\begingroup$ 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

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 $v = \frac{1}{25}w = \frac{1}{25}(40\cos(60), 40\sin(60)) = (1.6\cos(60), 1.6\sin(60))$

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.