Question

Find the foot of the perpendicular from the point (0, 2, 3) on the line `(-x - 3)/-5 = (1 - y)/-2 = (3z + 12)/9` and hence find the length of the perpendicular.

Answer 1

The given equation of the line is:

`(-x - 3)/-5 = (1 - y)/-2 = (3z + 12)/9`

First, we rewrite the given line equation in the standard symmetrical form `(x - x_1)/a = (y - y_1)/b = (z - z_1)/c`

`(-x - 3)/-5 = (x + 3)/5`

`(- (y - 1))/-2 = (y - 1)/2`

`(3z + 12)/9 = (z + 4)/3`

So, the line equation is:

`(x + 3)/5 = (y - 1)/2 = (z + 4)/3` = λ

Any point Q on this line can be expressed in terms of λ:

x = 5λ − 3

y = 2λ + 1

z = 3λ − 4

Thus, Q = (5λ − 3, 2λ + 1, 3λ − 4)

Let P(0, 2, 3) be the given point. If Q is the foot of the perpendicular, then the line PQ must be perpendicular to the given line.

Direction Ratios of PQ:

(5λ − 3 − 0, 2λ + 1 − 2, 3λ − 4  − 3) = (5λ − 3, 2λ − 1, 3λ − 7)

Direction Ratios of the given line: (5, 2, 3)

Since PQ ⊥ line, the sum of the products of their DRs is zero (a₁a₂ + b₁b₂ + c₁c₂ = 0):

5(5λ − 3) + 2(2λ − 1) + 3(3λ − 7) = 0

25λ − 15 + 4λ − 2 + 9λ − 21 = 0

38λ − 38 = 0

λ = 1

Substituting λ = 1 back into the coordinates of Q:

x = 5λ − 3

= 5(1) − 3

= 2

y = 2λ + 1

= 2(1) + 1

= 3

z = 3λ − 4

= 3(1) − 4

= −1

Foot of the perpendicular: Q(2, 3, −1)

The length of the perpendicular is the distance between P(0, 2, 3) and Q(2, 3, −1):

PQ = `sqrt((2 - 0)^2 + (3 - 2)^2 + (-1 - 3)^2)`

= `sqrt(2^2 + 1^1 + (-4)^2)`

= `sqrt(4 + 1 + 16)`

= `sqrt21`

Length of the perpendicular: `sqrt21` units