# Find the components of a vector?

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 :

• 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) ... – Matti P. Apr 12 at 7:12
• So I should use the formula (a, b) = k*(c ,d) ? but that is not enough right ? – Jonathcraft Apr 12 at 7:34
• 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. – 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 $$v = \frac{1}{25}w = \frac{1}{25}(40\cos(60), 40\sin(60)) = (1.6\cos(60), 1.6\sin(60))$$

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