How to do a C4 Vectors exam question

Look at part f) of the question In the second post of this thread. And a longer method is given below it that can be used if you don't know the angle.

If you have a different question that you are struggling with then please post it here.

Yes. In principle, the shortest distance between a line and a point is the perpendicular distance from the point to the line. If you can express the vector from the point to the line in its general form, then dot it with the direction of the line, and make this product equal 0 (due to perpendicularity), you should be able to find the parameter on the line where the shortest distance is achieved, thus have a specific vector from the point to the line. Then shortest distance is the magnitude of this vector.

I would highly recommend watching the video I've linked below (you may have to copy and paste the link into your web browser as my phone won't let me do it properly)

http://www.examsolutions.net/tutorials/shortest-distance-of-a-point-to-a-line/?level=A-Level&board=Edexcel&module=C4&topic=1567

So I'm stuck on part d of the Edexcel C4 January 2008 Paper (Question 6).

Can someone tell me how the vector equation of line $l_2$ was found? I don't understand

You are given the direction vector (a vector parallel to the line) and a point that the line passes through so you can form an equation of the line.

This is standard stuff that I think you know how to do...?

...

Yes you know that the origin lies on the line and the origin has position vector $\begin{pmatrix} 0\\ 0\\0 \end{pmatrix}$ so that is your $\mathbf{a}$ in the

$\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}$ formula.

The differentiation used in context question at the end is particularly rather difficult but fun to understand.

Ah that question was nasty. Arguably the trickiest part that always catches students out is where students use $k$ in their working as the constant of proportionality but the $k$ in the question is different.

Also "water is leaking out of the whole at 400" - I think a lot of students would set $\frac{dh}{dt}=400$.

But I'm glad you found it fun

Is there a way of identifying whether the angle between two vectors (of unknown directions) give an acute angle?

Obviously I can spot from the answer given when using $cos \theta = \frac{a . b}{ |a| |b|}$ that if $\theta > 90^{\circ}$ then it is clearly an obtuse angle. But is there an indication of whether it will be acute or obtuse beforehand?

Or should I just get used to using the mod version to always give me the acute angle?aka...

Assuming $\mathbf{a}$ and $\mathbf{b}$ are pointing towards/away from each other:

$\theta$ is obtuse implies that $\cos \theta$ is negative which implies that the numerator of $\frac{\mathbf{a}\cdot \mathbf{b}}{ |\mathbf{a}| |\mathbf{b}|}$ must be negative (since the denominator is always positive).

So the angle between $\mathbf{a}$ and $\mathbf{b}$ is obtuse when $\mathbf{a}\cdot \mathbf{b}$ is negative. I don't know why you would need to check this though in a standard exam question.

You could try and imagine the vectors in 3D space (can be hard) but again I don't see the point in this and it will waste time.

I've never found the modulus formula that useful. You may as well just find what the actual angle between two vectors pointing towards/away from each other is and show this on your diagram. And if they ask for the acute angle between the two vectors and yours is obtuse then they must be looking for the other angle between the two vectors so you just subtract from 180.

The shortest distance from A to OB is the length of the line that starts from A and meets OB at a right-angle. If you draw a diagram then you should see that you've formed a right-angled triangle where you have one of the angles in the triangle (well you have the cosine of that angle) and you can work out one of the lengths in the triangle then you can use SOHCAHTOA to find the shortest distance,

I've explained a very similar question here that you might find useful (part f).

Question though - For the area of the parallelogram (part e), why cant you do |AB| * |AD|

The area of a parallelogram is base x perpendicular height. Multiplying two of the adjacent lengths won't work since they're not perpendicular.

https://www.youtube.com/watch?v=gKRDtUsvBuA

So this is wrong?

That's the vector product of two vectors which is a further maths topic.

You wrote |AB| * |AD|

which I assumed meant the product of the magnitudes of AB and AD.

But if you do the magnitude of AB and AD it gives the same value in the case of the video.

But anyway - What is the best way to work out the area of the parallelogram? (C4 wise) - The way you did it? Using the area of the triangle * 2, with the angle ?