Question

Find the point of intersection of the line `vecr = (3hati + hatk) + μ(hati + hatj + hatk)` and the line through (2, –1, 1) parallel to the z-axis. How far is this point from the z-axis?

Answer 1

Let Point A be (2, −1, 1).

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

Equation of line is `vecr = (3hati + hatk) + μ(hati + hatj + hatk)`

Now, point B is the point of intersection of line `vecr = (3hati + hatk) + μ(hati + hatj + hatk)` and the line through (2, −1, 1). parallel to the z-axis.

Finding Point B:

Since point B lies on line `vecr`.

Now,

`vecr = (3hati + hatk) + μ(hati + hatj + hatk)`

`vecr = 3hati + hatk + μhati + μhatj + μhatk`

`vecr = (3 + μ)hati + μhatj + (1 + μ) hatk`

So, x = 3 + μ

y = μ

z = 1 + μ

∴ B = (3 + μ, μ, 1 + μ)

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, 1) and B(3 + μ, μ, 1 + μ)

Direction ratios of AB:

⇒ 3 + μ − 2

= 1 + μ

⇒ μ − (−1)

= μ + 1

⇒ 1 + μ − 1

= μ

Direction cosines of AB = `(1 + μ)/(AB), (1 + μ)/(AB), μ/(AB)`

Since AB and the z-axis are parallel.

Both sets of direction cosines are equal.

Equating x-component:

`(1 + μ)/(AB) = 0`

1 + μ = 0

μ = −1

Thus, point B becomes:

x = 3 + μ

= 3 − 1

= 2

y = μ = −1

z = 1 + μ

= 1 + (−1)

= 0

∴ B = (2, −1, 0)

Thus, the required point B is (2, −1, 0).

Now, we need to find distance between B(2, −1, 0) and z-axis.

Distance of B (x, y, z) and z-axis = `sqrt(x^2 + y^2)`

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

= `sqrt(4 + 1)`

= `sqrt5` units