Question

Find the length of the perpendicular (2, –3, 1) to the line `(x + 1)/(2) = (y - 3)/(3) = (z + 1)/(-1)`.

Answer 1

Let PM be the perpendicular drawn from the point P(2, –3, 1) to the line `(x + 1)/(2) = (y - 3)/(3) = (z + 1)/(-1) = λ`   ...(Say)

The coordinates of any point on the line are given by

x = – 1 + 2λ, y = 3 + 3λ, z = – 1 – λ
Let the coordinates of M be
(–1 + 2λ, 3 + 3λ, –1 – λ)       ...(1)

The direction ratios of PM are
–1 + 2λ – 2, 3 + 3λ + 3, –1 – λ –1
i.e. 2λ – 3, 3λ + 6, – λ – 2

The direction ratios of the given line are 2, 3, –1.
Since PM is perpendicular to the given line, we get
2(2λ – 3) + 3(3λ + 6) – 1(– λ – 2) = 0
∴ 4λ – 6 + 9λ + 18 + λ + 2 = 0
∴ 14λ + 14 = 0
∴ λ = – 1.
Put λ = – 1 in (1), the coordinates of M are
(– 1 – 2, 3 – 3, – 1 + 1) i.e. (– 3, 0, 0).
∴ length of perpendicular from P to the given line
= PM

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

= `sqrt((25 + 9 + 1)`

= `sqrt(35) "units"`.

Answer 2

We know that the perpendicular distance from the point

`"P"|barα|  "to the line"  bar"r" = bar"a" + λbar"b"` is given by

`sqrt(|barα - bar"a"|^2 - [((barα - bara).barb)/[|bar"b"|)]^2`                 ...(1)

Here, `barα = 2hat"i" - 3hat"j" + hat"k", bar"a" = -hat"i" + 3hat"j" - hat"k", bar"b" = 2hat"i" + 3hat"j" - hat"k"`

∴ `barα - bar"a" = (2hat"i" - 3hat"j" + hat"k") - (-hat"i" + 3hat"j" - hat"k")`

= `3hat"i" - 6hat"j" + 2hat"k"`

∴ `|barα - bar"a"|^2` = 3² + (– 6)² + 2² = 9 + 36 + 4 = 49

Also, `(barα - bar"a").bar"b" = (3hat"i" - 6hat"j" + 2hat"k").(2hat"i" + 3hat"j" - hat"k")`

= (3)(2) + (– 6)(3) + (2)(– 1)
= 6 – 18 – 2
= – 14

`|bar"b"| = sqrt(2^2 + 3^2 + (-1)^2) = sqrt(14)`
Substitutng these values in (1), we get
length of perpendicular from P to given line
= PM
= `sqrt(49 - ((-14)/sqrt(14))^2`

= `sqrt(49 - 14)`

= `sqrt(35)  "units"`.