Question

Find the perpendicular distance of the point (1, 0, 0) from the line  \[\frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 10}{8}.\] Also, find the coordinates of the foot of the perpendicular and the equation of the perpendicular.

Answer 1

Given: Point P(1, 0, 0) and equation of line `(x - 1)/2 = (y + 1)/-3 = (z + 10)/8`

Let, PQ be the perpendicular drawn from P to given line whose endpoint/foot is Q point.

Thus, to find Distance PQ we have to first find coordinates of Q

`(x - 1)/2 = (y + 1)/-3 = (z + 10)/8` = λ(let)

⇒ x = 2λ + 1, y = – 3λ – 1, z = 8λ – 10

Therefore, coordinates of Q(2λ + 1, – 3λ – 1,8λ – 10)

Now as we know (TIP) ‘if two points A(x₁, y₁, z₁) and B(x₂, y₂, z₂) on a line, then its direction ratios are proportional to (x₂ – x₁, y₂ – y₁, z₂ – z₁)'

Hence, Direction ratio of PQ is

= (2λ + 1 – 1), ( – 3λ – 1 – 0), (8λ – 10 – 0)

= (2λ), ( – 3λ – 1), (8λ – 10)

and by comparing with given line equation, direction ratios of the given line are

= (2, −3, 8)

Since PQ is perpendicular to given line, therefore by “condition of perpendicularity.”

a₁a₂ + b₁b₂ + c₁c₂ = 0; where a terms and b terms are direction ratio of lines which are perpendicular to each other.

⇒ 2(2λ) + ( – 3)( – 3λ – 1) + 8(8λ – 10) = 0

⇒ 4λ + 9λ + 3 + 64λ – 80 = 0

⇒ 77λ – 77 = 0

⇒ λ = 1

Therefore coordinates of Q

i.e. Foot of perpendicular

By putting the value of λ in Q coordinate equation, we get

= Q (2(1) + 1, –3(1) – 1, 8(1) – 10)

= Q (3, –4, –2)

Now,

Distance between PQ

Distance between two points A(x₁, y₁, z₁) and B(x₂, y₂, z₂) is given by

= `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2)`

= `sqrt((1 - 3)^2 + (0 + 4)^2 + (-2 - 0)^2)`

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

= `sqrt(4 = 16 + 4)`

= `sqrt24`

= 2√6 unit