Question

A school is preparing the stage for its annual day function. They want to place a hanging mic and a hanging light on the stage.

• They decide to position the mic at the point (3,2,1) such that it is equidistant from a plain backdrop and the hanging light, as shown below.
  
• The equation of the surface of the plain backdrop is 2x − y + z + 1 = 0

(a) Find the distance between the mic and the plain backdrop.

(b) Calculate the coordinates of the position of the hanging light.

Answer 1

Given:

Mic Position (M): (3, 2, 1)

Equation of Backdrop (Plane): 2x − y + z + 1 = 0

Condition: Mic (M) is the midpoint between the backdrop point (P) and the light (L).

(a) Distance between the mic and the plain backdrop

The perpendicular distance (d) from the point (x₁, y₁, z₁) to the plane is given by:

`d = |ax_1 + by_1 + cz_1 + d|/sqrt(a^2 + b^2 + c^2)` 

Substituting (x₁, y₁, z₁) = (3, 2, 1) and the plane equation 2x − y + z + 1 = 0

d = `|2(3) − 1(2) + 1(1) + 1|/sqrt(2^2 + (−1)^2 + 1^2)`

d = `|6 − 2 + 1 + 1|/sqrt(4 + 1 + 1)`

d = `6/sqrt6`

d = `sqrt6` units

(b) Coordinates of the position of the hanging light

Point P (Foot of the Perpendicular on the Backdrop)

The equation of the line passing through M(3, 2, 1) and perpendicular to the plane is:

`(x − 3)/2 = (y − 2)/(−1) = (z − 1)/1 = λ`

General coordinates for point P.x = 2λ + 3, y = −λ + 2, z = λ + 1

Substituting these into the plane equation 2x − y + z + 1 = 0

2(2λ + 3) − (−λ + 2) + (λ + 1) + 1 = 0

4λ + 6 + λ − 2 + λ + 1 + 1 = 0

6λ + 6 = 0

⇒ λ = −1

Now, substitute λ = −1 back into the coordinates for P.

P = (2(−1) + 3, − (−1) + 2, (−1) + 1) = (1, 3, 0)

Find Light (L) coordinates using the Midpoint Formula

Let the coordinates of light L be (x, y, z). Since M(3, 2, 1) is the midpoint of PL:

`3 = (1 + x)/2 ⇒ 6 = 1 + x ⇒ x = 5` 

`2 = (3 + y)/2 ⇒ 4 = 3 + y ⇒ y = 1`

`1 = (0 + z)/2 ⇒ 2 = 0 + z ⇒ z = 2`

The coordinates of the hanging light are (5, 1, 2).