The Student Room Group

basic vector question

question and answer is found in the attachment

i understand everything up until the part i highlighted in red. how is the distance from the origin and the line calculated?

many thanks in advance :smile:
Hi there,

While you're waiting for an answer, did you know we have 300,000 study resources that could answer your question in TSR's Learn together section?

We have everything from Teacher Marked Essays to Mindmaps and Quizzes to help you with your work. Take a look around.

If you're stuck on how to get started, try creating some resources. It's free to do and can help breakdown tough topics into manageable chunks. Get creating now.

Thanks!

Not sure what all of this is about? Head here to find out more.
Reply 2
Original post by shartos maximus
question and answer is found in the attachment

i understand everything up until the part i highlighted in red. how is the distance from the origin and the line calculated?

many thanks in advance :smile:


To calculate the distance of a point from a line, you need for
the equation of the line, and the coordinates of the point.

1. First of all, we arrange the equation of the line to zero.
Let the form of equation be Ax+By+C=0

This equation is true (is 0) for all points (x,y), whiches are on the line,
C=-Ax0-By0, where (x0,y0)
is the coordinates of a given P0 point on the line.

So the equation in original form:
A(x-x0)+B(y-y0)=0
Where (x,y) the coordinates of a P point running on the line

From this follows that this equation states that the dot product of the
normal vector (A,B), and the PqP vector (x-x0,y-y0) is zero, becauise they are mutually perpendicular vectors.

2. Substituting the coordinates of so P point, which is not on the line into x and y,
the equation gives the dot product of P0P vector and the normal vector
which mean the product of length of the normal vector and length of the
P0P vector projected on the normal vector.

So the normal vector length is the unit then the pjojected length
geometrically is the distance of P from the line

3. So Divide the equation of the line by V(A^2+B^2)

4. A/ V(A^2+B^2)x+B/V(A^2+B^2)y+C/V(A^2+B^2)=d

in this form the equation gives the signed distance of P fron the line

Quick Reply