mind blank on part 13 f (1 Viewer)

[amdspotter]

any help is appreciated
for reference i have attached my answers for q13 a b c d e:

 

[yanujw]

Firstly, vector diagrams always help. Here is one for this situation

The blue line is line l
The projection is the vector from A to the intersection of the purple and blue lines in the bottom right
The purple line is the 'perpendicular distance' we aim to solve for
The black line is vector AP
The projection is perpendicular to the vector that it projects onto (line l). So the intersection of the blue and purple lines is a right angle.
You now have a right-angled triangle. Solving by pythagoras' theorem gives the answer of

 

[Run hard@thehsc]

@yanujw I got that too!

 

[5uckerberg]

View attachment 34593
any help is appreciated
for reference i have attached my answers for q13 a b c d e:

With question 13f another approach while awkward at first but quite elegant if you are asking me. Using what we know from Q13a where the gradient is .

Right, since we know that with the perpendicular gradient formula it goes like this so thus by default, the gradient is 3. Now we will have something along the lines of where b is the intercept and for more simplicity sub yes you will get . The next step will be to do simultaneous equations for


Well, here

Then find the length by doing