Question

Find a point P on the line `(x + 5)/1 = (y + 3)/4 = (z - 6)/-9` such that its distance from point Q(2, 4, – 1) is 7 units. Also, find the equation of line joining P and Q.

Answer 1

Given: `(x + 5)/1 = (y + 3)/4 = (z - 6)/-9 = λ`

x = λ − 5, y = 4λ − 3, z = −9λ + 6

P(λ − 5, 4λ − 3, −9λ + 6)

Given, PQ = 7

x₁ = λ − 5, y₁ = 4λ − 3, z₁ = −9λ + 6

x₂ = 2, y₂ = 4,  z₂ = −1

So, `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 + z_1)^2) = 7`

`sqrt((λ - 5 - 2)^2 + (4λ - 3 - 4)^2 + (-9λ + 6 + 1)^2) = 7`

`sqrt((λ - 7)^2 + (4λ - 7)^2 + (-9λ + 7)^2) = 7`

Squaring on both sides

λ² + 49 − 14λ+ 16λ² + 49 − 56λ + 81λ² + 49 − 126λ = 49

98λ² − 196λ + 98 = 0

98(λ² − 2λ + 1) = 0

(λ − 1)² = 0

λ = 1

∴ Cooridinates of p[1 − 5, 4(1) − 3, −9(1) + 6]

= −4, 1, −3

Equation of line PQ is 

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

PQ: `(x + 4)/6 = (y - 1)/3 = (z + 3)/2`