Maths C4 - Vectors... Help???

So I've made a little start on this chapter but have found myself stuck already!!

How do I do the last bit of this Example question after I have equated the coefficients?? I don't really understand how it goes from the 3rd from bottom line to the 2nd from bottom line of the workings

You equate coefficients in the same as as you'd equate coefficients of something like $x^2+2x+1=Ax^2+Bx+C$ except here you are comparing the coefficients of the direction vectors $\mathbf{a}$ and $\mathbf{b}$

Yeah I think I understand the part where I compare coefficients but I dont really know the eliminating part which gives the 2nd from bottom line.

Is it simply a case of using simultaneous equations where...
$-\land = (k-1)$ times both sides by -1 to give... $\land = (1-k)$

Sub this into... $(3k+1) = 4 \land$
which gives... $(3k+1) = 4(1-k)$

then solve for k from there?


Sorry I'm not sure what the correct symbol is so I have used $\land$ instead

Yeah thats what is being done here. Also the symbol is lambda.


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If $k-1=-\lambda$ then $4\lambda = -4(k-1)$ by just multiplying both sides by 4.
Now we have also that $3k+1=4\lambda$ so you can substitute $4\lambda$ with $-4(k-1)$ which gives $3k+1=-4(k-1)$.

Thank you!!

So in a way, lambda $\lambda$ is used to describe scalars that are parallel and proportional to each other, right?

Lambda is just a scalar variable which scales vectors by changing their magnitudes. It doesnt make sense when you say "used to describe scalars that are parallel and proportional" because scalars have no direction so how can they be parallel...?

Yeah your explanation makes more sense than mine! Thank you!! I had trouble putting it into words but had a brief understanding.

So, lambda is a scalar variable which scales vectors by changing their magnitude. And the vectors have to be parallel too for them to have a related lambda? Sorry if I sound silly right now!

Yeah your explanation makes more sense than mine! Thank you!! I had trouble putting it into words but had a brief understanding.

So, lambda is a scalar variable which scales vectors by changing their magnitude. And the vectors have to be parallel too for them to have a related lambda? Sorry if I sound silly right now!

No, the scalar has nothing to do with dependency on two vectors being parallel or not, any vector can have its size changed by a vector regardless whether it is parallel to another vector or not. The only use for scalars here is that they change the magnitude, as previously mentioned, so you can use this to travel from some point A to some point B on whatever vector v by adding on $\lambda \mathbf{v}$ onto the point A.

$\lambda$ isn't anything special - it just represents a scalar variable.

So if you have a vector $\mathbf{a}$ then $\lambda \mathbf{a}$ could be $2 \mathbf{a}$ or $-3.1\boldsymbol{a}$ etc, depending on the value of $\lambda$.

You can use any letter really as long as you don't underline it so it represents a scalar. $\lambda$ and $\mu$ are quite common.

It is the case that a scalar $\lambda$ multiplied by a vector $\mathbf{a}$ will give you a new vector $\lambda \mathbf{a}$, which is parallel to $\mathbf{a}$ but whose magnitude has changed by a factor of $\lambda$.

Wait so... $\lambda a$ will give a new vector that is parallel to the original vector $a$ ??

Yes e.g. $2\mathbf{a}$ is parallel to $\mathbf{a}$:

Stupid question of the day...
What is a position vector?? Are these vectors that are from O the origin?



Like with this example. why is OP the position vector of P?

Can someone explain perpendicular and parallel vectors to me?...



I've read this little section multiple times but I don't think I completely understand. Obviously I know that if there are two lines perpendicular to each other that there will be an angle of 90 degrees, but the rest of it I'm not so sure about

Can someone explain perpendicular and parallel vectors to me?...



I've read this little section multiple times but I don't think I completely understand. Obviously I know that if there are two lines perpendicular to each other that there will be an angle of 90 degrees, but the rest of it I'm not so sure about

Try writing out the dot product formula for parallel lines $\theta = 0$ and perpendicular lines $\theta = 90$.

Then use the fact that $\cos (0) = 1$ and $\cos (90) = 0$.

Please post your working if you're still unsure.

Firstly, are you happy about where and how we get $\mathbf{a}\cdot \mathbf{b}=\lvert \mathbf{a}\lvert \lvert \mathbf{b} \lvert \cos(\theta)$ ?

Secondly, if two vectors are parallel then the angle between them is always 0 so plugging this into the equation gives you the shown relation.

Dunno but maybe watching this video can help since that relation is derived towards the end https://www.youtube.com/watch?v=WNuIhXo39_k

Thank you @notnek!! I understand perpendicular vectors a lot more now! But I'm still a bit "iffy" with parallel vectors.

So for parallel vectors there will be no angle between them so if then $cos(0 ^\circ ) =1$

because using the scalar product...
$cos(\theta) = \frac{a.b}{|a||b|}$

$cos(0^\circ) = 1$

This shows that...
$\frac{a.b}{|a||b|} = 1$ , when $\theta = 0^\circ$

Meaning...
$a.b = |a||b|$