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My question is regarding the following problem:

Relative to a fixed origin, O, the line l has the equation

r = (i + pj − 5k) + λ(3i − j + qk),

where p and q are constants and λ is a scalar parameter.

Given that the point A with coordinates (−5, 9, −9) lies on l,

(a) find the values of p and q,

(b) show that the point B with coordinates (25, −1, 11) also lies on l.

The point C lies on l and is such that OC is perpendicular to l.

(c) Find the coordinates of C.

(d) Find the ratio AC : CB

I have no problem with parts a to c (for a the answer is p=7, q=2, and for c the answer is (4, 6, −3)). However, for part (d) the method which I used was simply finding the magnitude of AC and CB, and then finding the ratio. However, the markscheme has used a far more elegant method, which I was wondering if someone could explain:

A : λ = −2, B : λ = 8, C : λ = 1 ∴ AC : CB = 3 : 7

Thank you for any assistance.

Relative to a fixed origin, O, the line l has the equation

r = (i + pj − 5k) + λ(3i − j + qk),

where p and q are constants and λ is a scalar parameter.

Given that the point A with coordinates (−5, 9, −9) lies on l,

(a) find the values of p and q,

(b) show that the point B with coordinates (25, −1, 11) also lies on l.

The point C lies on l and is such that OC is perpendicular to l.

(c) Find the coordinates of C.

(d) Find the ratio AC : CB

I have no problem with parts a to c (for a the answer is p=7, q=2, and for c the answer is (4, 6, −3)). However, for part (d) the method which I used was simply finding the magnitude of AC and CB, and then finding the ratio. However, the markscheme has used a far more elegant method, which I was wondering if someone could explain:

A : λ = −2, B : λ = 8, C : λ = 1 ∴ AC : CB = 3 : 7

Thank you for any assistance.

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