Question

Let the line L be the projection of the line: `(x - 1)/2 = ("y" - 3)/1 = ("z" - 4)/2` in the plane x – 2y – z = 3. If d is the distance of the point (0, 0, 6) from L, then d² is equal to ______.

विकल्प

• 24

• 25

• 26

• 27

Answer 1

Let the line L be the projection of the line: `(x - 1)/2 = ("y" - 3)/1 = ("z" - 4)/2` in the plane x – 2y – z = 3. If d is the distance of the point (0, 0, 6) from L, then d² is equal to 26.

Explanation:

A₁, B₁ ⇒ foot of ⊥ A, B

`(α - 1)/1 = (β - 3)/(-2) = (ϒ - 4)/(-1) = (-(1 - 6 - 4 - 3))/6` = 2

⇒ A₁ = (3, –1, 2)

`(α - 3)/1 = (β - 4)/(-2) = (ϒ - 6)/(-1) = (-(3 - 8 - 6 - 3))/6 = 7/3`

⇒ B₁ = `(16/3, (-2)/3, 11/3)`

Drs of A₁B₁: `16/3 - 3, (-2)/3 + 1, 11/3 - 2`

⇒ `7/3, 1/3, 5/3`

∴ L: `(x - 3)/(7/3) = ("y" + 1)/(1/3) = ("z" - 2)/(5/3)` = ℓ

`(7/3ℓ + 3, ℓ/3 - 1, 5/3ℓ + 2)` = B

Drs of AB: `7/3ℓ + 3, ℓ/3 - 1, 5/3ℓ - 4`

`bar("AB") ⊥ "L"` ⇒ `7/3(7/3ℓ + 3) + 1/3(ℓ/3 - 1) + 5/3(5/3ℓ - 4)` = 0

`(49/9 + 1/9 + 25/9)l + 21/3 - 1/3 - 20/3` = 0

`75/9ℓ` = 0 ⇒ ℓ = 0. So the point B = (3, –1, 2)

∴ Distance, AB = d

= `sqrt((3 - 0)^2 + (-1 - 0)^2 + (2 - 6)^2)`

`sqrt(9 + 1 + 16) = sqrt(26)`

⇒ d² = 26