Question

Find the distance of the point (2, –1, 3) from the line `vecr = (2hati - hatj + 2hatk) + μ(3hati + 6hatj + 2hatk)` measured parallel to the z-axis.

Answer 1

Let Point A be (2, –1, 3).

Let point B be the point on line `hatr`, such that AB is parallel to the z-axis.

Equation of line is:

`vecr = (2hati - hatj + 2hatk) + μ(3hati + 6hatj + 2hatk)`

We need to find Distance AB.

To find AB, we need to find point B Finding Point B.

Finding Point B:

Since point B lies on line `vecr`

Now,

`vecr = (2hati - hatj + 2hatk) + μ(3hati + 6hatj + 2hatk)`

`vecr = 2hati - hatj + 2hatk + 3μhati + 6μhatj + 2μhatk`

`vecr = (2 + 3μ)hati + (1 + 6μ)hatj + (2 + 2μ)hatk`

So, x = 2 + 3μ

y = −1 + 6μ

z = 2 + 2μ

∴ B = (3μ + 2, 6μ − 1, 2μ + 2)

Since AB is parallel to the z-axis.

Their direction cosines would be equal.

Direction cosines of the z-axis are:

I = cos 90° = 0

m = cos 90° = 0

n = cos 0° = 1

∴ Direction cosines of the z-axis are 0, 0, 1.

Direction cosines of AB:

For A(2, –1, 3) and B = (3μ + 2, 6μ − 1, 2μ + 2)

Direction ratios of AB:

= 3μ + 2 – 2, 6μ − 1 − (−1), 2μ + 2 − 3

= 3μ, 6μ, 2μ − 1

Direction cosines of AB = `(3μ)/(AB), (6μ)/(AB), (2μ - 1)/(AB)`

Since AB and the z-axis are parallel.

Both sets of direction cosines are equal.

Equating x-component:

`(3μ)/(AB)` = 0

3μ = 0

u = 0

Thus, point B becomes:

x = 2 + 3μ

= 2 + 2(0)

= 2 + 0

= 2

y = −1 + 6μ

= −1 + 6(0)

= −1 + 0

= −1

z = 2 + 2μ

= 2 + 2(0)

= 2 + 0

= 2

∴ B = (2, −1, 2)

Thus, the distance between A(2, −1, 3) and B (2, −1, 2):

AB = `sqrt((2 - 2)^2 + [(-1) - (-1)]^2 + (2 - 3)^2)`

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

= `sqrt1`

= 1 unit

Thus, the required distance is 1 unit.