Finding vector coordinates help

[nokia8650]

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.

 

[HallsofIvy]

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.

I take it you mean that in parts (a), (b), and (c), you have determined that the corresponding $\lambda$ values for A, B, and C are -2, 8, and 1 respectively.

Write the equation of the line as P+ tQ. Then A= P-2Q, B= P+ 8Q, C= P+ Q. A-C is the vector (-2-(1))Q= -3Q and B- C is the vector (8- 1)Q= 7Q. Since the lengths AC and BC are 3|Q| and 7|Q| respectively the ratio of their lengths is 3/7.

 

[nokia8650]

ahh yes, of course! Thank you!