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difficult fp3 vectors

can anyone help me out with 6d and 7c http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/june2010-qp/6669_01_que_20100628.pdf

im not sure how to turn the cartesian into vector equation and no idea how to work out perpendicular distance :O

thanks :smile:
Reply 1
Avatar for BabyMaths
BabyMaths
Probably not the method they have in mind but, you can easily find two points on l1.

Apply the transformation to get two points on l2.

Write down the equation of l2 in vector form and convert to cartesian equations.
Reply 2
Avatar for BabyMaths
BabyMaths
If you want to change x−ab=x−cd=z−ef\frac{x-a}{b}=\frac{x-c}{d}=\frac{z-e}{f} to a vector equation set them all equal to t.

x−ab=t⇒x=a+bt\frac{x-a}{b}=t\Rightarrow x=a+bt

y−cd=t⇒y=c+dt\frac{y-c}{d}=t\Rightarrow y=c+dt

z−ef=t⇒z=e+ft\frac{z-e}{f}=t\Rightarrow z=e+ft
Reply 3
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number23
OP
15
Original post by BabyMaths
If you want to change x−ab=x−cd=z−ef\frac{x-a}{b}=\frac{x-c}{d}=\frac{z-e}{f} to a vector equation set them all equal to t.

x−ab=t⇒x=a+bt\frac{x-a}{b}=t\Rightarrow x=a+bt

y−cd=t⇒y=c+dt\frac{y-c}{d}=t\Rightarrow y=c+dt

z−ef=t⇒z=e+ft\frac{z-e}{f}=t\Rightarrow z=e+ft


so those equations for x,y and z are the vector coordinates?
Reply 4
Avatar for BabyMaths
BabyMaths
Original post by number23
so those equations for x,y and z are the vector coordinates?


How can equations be coordinates?

To make it clearer..

(xyz)=(ace)+t(bdf)\begin{pmatrix}x \\ y\\ z \end{pmatrix}= \begin{pmatrix}a \\ c\\e \end{pmatrix}+ t \begin{pmatrix}b \\ d \\ f \end{pmatrix}
Reply 5
Avatar for number23
number23
OP
15
Original post by BabyMaths
How can equations be coordinates?

To make it clearer..

(xyz)=(ace)+t(bdf)\begin{pmatrix}x \\ y\\ z \end{pmatrix}= \begin{pmatrix}a \\ c\\e \end{pmatrix}+ t \begin{pmatrix}b \\ d \\ f \end{pmatrix}


yeah this us what i meant.. the x coordinate of a point in the line is given by an equation and so forth

thanks for the help
Reply 6
Avatar for number23
number23
OP
15
Original post by BabyMaths
Probably not the method they have in mind but, you can easily find two points on l1.

Apply the transformation to get two points on l2.

Write down the equation of l2 in vector form and convert to cartesian equations.


Original post by BabyMaths
If you want to change x−ab=x−cd=z−ef\frac{x-a}{b}=\frac{x-c}{d}=\frac{z-e}{f} to a vector equation set them all equal to t.

x−ab=t⇒x=a+bt\frac{x-a}{b}=t\Rightarrow x=a+bt

y−cd=t⇒y=c+dt\frac{y-c}{d}=t\Rightarrow y=c+dt

z−ef=t⇒z=e+ft\frac{z-e}{f}=t\Rightarrow z=e+ft


thanks, also once youve got these equations for x, y etc. how do you find the vector equation for the line?

also, could anyone help me with this one:

the cartersian equation of a line is: (x−x1)/l=(y−y1)/m=(z−z1)/n=lambda (x-x_1)/l=(y-y_1)/m=(z-z_1)/ n=lambda

my book says the line has direction ratios l : m : n ... what does this mean? thanks :smile:

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