C4 - Vector parallel to the Z axis, why do I take it as (0,0,1)?

Find a vector equation or the line which is parallel to the z-axis and passes through the point (4,-3,8).

I know the third coordinate is the z value, and looking at the mark scheme, I can see that for a, you take a to be (0,0,1). Why is this though? Why do you know that the parallel to z axis bit makes the z value 1 and the x and y values 0?

Also, for the equation r=a+tb, how do you know which is a and which is b in this case? The mark scheme says the (0,0,1) is the a value, but how do you know to take (0,0,1) to be a and not taking it to be b?

Find a vector equation or the line which is parallel to the z-axis and passes through the point (4,-3,8).

I know the third coordinate is the z value, and looking at the mark scheme, I can see that for a, you take a to be (0,0,1). Why is this though? Why do you know that the parallel to z axis bit makes the z value 1 and the x and y values 0?

Also, for the equation r=a+tb, how do you know which is a and which is b in this case? The mark scheme says the (0,0,1) is the a value, but how do you know to take (0,0,1) to be a and not taking it to be b?

Because you (should) know that the x-axis has direction $1 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k}$, then the y-axis has direction $0 \mathbf{i} + 1 \mathbf{j} + 0 \mathbf{k}$, and so the z-axis has direction $0 \mathbf{i} + 0 \mathbf{j} + 1 \mathbf{k}$.

These are all linearly independent vectors, which gives the perpendicular axes of the system. Hence, for a vector parallel to the z-axis, you can just pick (0,0,1), which is the same direction as the z-axis itself.

The markscheme should say that a=(4,-3,8) and b=(0,0,1)

Ok so remember the vector equation is in the form r=a+tb where is a particular position vector (in this case (4,-3,8)), t is a scalar constant and b is a directional vector. So essentially you need a line in the 3D plane that passes through (4,-3,8) so therefore this point must lie on the vector line. The easiest way to ensure this is let a be the position vector (4,-3,8) Note I really should be writing vectors with vertical columns but Idk how to .

Now the other part of the equation is making the line parallel to the z axis- that must mean the directional vector parallel to the z axis must be in the form (0,0,m) where m is essentially any real number. The reason why the x and y values are 0 is to ensure the vector is parallel to thee z axis otherwise it wouldn't be- imagine the 3D coordinate axes and having any line parallel to the z axis and finding 2 points on that line - the two points will definitely have the same x and y values and the only thing that will be different will be there z values so if you imagine finding the vector that connects those two lines it will be (p-p, q-q, s-u) which means that x and y values become 0 and only the z value has a non zero number. The mark scheme has let m=1 for simplicity but technically it can be any real number as the scalar constant t can take any real value.

therefore one possible vector equation for this line is r= (4,-3,8) + t(0,0,1)

Hope this helped

How do you know which is a and which is b though? And why should a=(4,-3,8) and not (0,0,1)?

Thank you. How can you tell which is a position vector and which is a directional vector? And so how can you tell that a should be (4,-3,8) and not the (0,0,1)?