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Edexcel C4 Vectors Question - Perpendicular/shortest distance from a point to a line

Reference: Edexcel C4 book, Mixed Ex 5K Q10 part (d).I have named a point C on the line, where BC is the perpendicular distance from B to the line. I have called C's coordinates (x, y, z). So, the vector BC (x-10, y+2, z-2) dotted with the directional vector of the line (3, -1, 1) must equal 0 (because they're perpendicular). When I worked this out, I found x=10, y=-2 and z=2. For some reason this gave me position vector of B, not the position vector of C. Can anyone tell me what I've done wrong?
You would need to dot BC with

OA+(3,1,1)\overrightarrow{OA}+ (3,-1,1)

As that's the equation of the line, what you have used is a line that is not parallel to l, it is a point from the origin to a point that l lies on.

It is easier however to use the sine rule from a right angled triangle formed as angle BCA, as you know angle CAB and the length AB and can therefore work out angle CBA.
(edited 7 years ago)
Original post by glass57
Reference: Edexcel C4 book, Mixed Ex 5K Q10 part (d).I have named a point C on the line, where BC is the perpendicular distance from B to the line. I have called C's coordinates (x, y, z). So, the vector BC (x-10, y+2, z-2) dotted with the directional vector of the line (3, -1, 1) must equal 0 (because they're perpendicular). When I worked this out, I found x=10, y=-2 and z=2. For some reason this gave me position vector of B, not the position vector of C. Can anyone tell me what I've done wrong?


Well I'm a bit confused on the highlighted part because you need to use the equation of the line (which I do not know) so I will say it is r=(1,2,3)+t(3,1,1)r=(1,2,3)+t(3,-1,1) and since C is on the line, the coordinates of C are C(1+3t,2t,3+t)C(1+3t, 2-t, 3+t).

Meaning BC (again, I dont know what B is so I will call it B(5,6,7) ) would be [(1+3t)-5, (2-t)-6, (3+t)-7]=(-4+3t, -4-t, -4+t) and NOW you can dot this with the direction vector (make sure it is head-to-head or tail-to-tail) and find t, then find C.
(edited 7 years ago)
it is always best to post the question ... either link to it or post a scan.

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