Question

The point A, with coordinates (0,a,b) lies on the line l1, which has equation:

Answer 1

a) The solution of the system is: (λ, a, b) = (- 5, - 11, - 26).

b) The coordinates of the point P are (x, y, z) = (- 16 / 3, - 43, 152 / 3).

c) Vectors AP and PB are collinear for their ends lie on the line. The ratio AP : PB is 83.248 : 73.491 (1.133 : 1).

How to analyze a line equation in vector form

a) In this problem we have a equation of the line in vector form, whose form is:

r = (5, 19, - 1) + λ · (1, 6, - 5)

Now we find the value of λ for the case r = (0, a, b):

(0, a, b) = (5, 19, - 1) + λ · (1, 6, - 5)

5 + λ = 0

19 + 6 · λ = a

- 1 + 5 · λ = b

The solution of the system is: (λ, a, b) = (- 5, - 11, - 26).

b) Then, the following condition must be fulfilled:

OP • PB = 0

(5 + λ, 19 + 6 · λ, - 1 + 5 · λ) • [(4, 13, 4) - (5 + λ, 19 + 6 · λ, - 1 - 5 · λ)] = 0

(5 + λ, 19 + 6 · λ, - 1 + 5 · λ) • (- 1 - λ, - 6 - 6 · λ, 5 + 5 · λ) = 0

(5 + λ) · (- 1 - λ) + (19 + 6 · λ) · (- 6 - 6 · λ) + (- 1 + 5 · λ) · (5 + 5 · λ) = 0

- (5 + λ) · (1 + λ) - 6 · (19 + 6 · λ) · (1 + λ) + 5 · (- 1 + 5 · λ) · (1 + λ) = 0

- (5 + 6 · λ + λ²) - 6 · (19 + 25 · λ + 6 · λ²) + 5 · (- 1 + 4 · λ + 5 · λ²) = 0

- 5 - 6 · λ - λ² - 114 - 150 · λ - 36 · λ² - 5 + 20 · λ + 25 · λ² = 0

- 12 · λ² - 136 · λ - 124 = 0

There are two solutions: λ₁ = - 1, λ₂ = - 31 / 3, whose points P are:

r₁ = (5, 19, - 1) + (- 1) · (1, 6, - 5)

r₁ = (4, 13, 4)

r₂ = (5, 19, - 1) + (- 31 / 3) · (1, 6, - 5)

r₂ = (- 16 / 3, - 43, 152 / 3)

Then, the coordinates of the point P are (x, y, z) = (- 16 / 3, - 43, 152 / 3).

c) Vectors AP and PB are collinear for their ends lie on the line. The ratio can be found by using the magnitudes of each vector:

Vector AP

AP = (- 16 / 3, - 43, 152 / 3) - (0, - 11, - 26)

AP = (- 16 / 3, - 32, 230 / 3)

AP = √(AP • AP)

AP ≈ 83.248

Vector PB

PB = (4, 13, 4) - (- 16 / 3, - 43, 152 / 3)

PB = (28 / 3, 56, - 140 / 3)

PB = √(PB • PB)

PB ≈ 73.491

The ratio AP : PB is 83.248 : 73.491 (1.133 : 1).

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