Plane Equations and distances

Question

 

 

Question 2

 

Answer 1

a).

The plane is αx + 3y + 2z = 9 and the point is P(3, 2 , - 3).

The plane contains the point P(3, 2, - 3).

Substitute the values of (x, y, z) = (3, 2, - 3).

α(3) + 3(2) + 2(- 3) = 9

3α + 6 - 6 = 9

α = 9/3

⇒ α = 3.

The value of α = 3.

b).

The new plane 3x + 3y + 2z = k is parallel to αx + 3y + 2z = 9 and contains the point D(3, 2, 2).

The new plane contains the point D(3, 2, 2).

Substitute the values of (x, y, z) = (3, 2, 2).

3(3) + 3(2) + 2(2) = k

9 + 6 + 4 = k

⇒ k = 19.

The value of k = 19.

c).

The two parallel planes are 3x + 3y + 2z = 9 and 3x + 3y + 2z = 19.

To find the distance between the two parallel planes choose a point in the first plane say (x₀, y₀, z₀) = (3, 2 , - 3).

From the second plane, we can determined that, a = 3, b = 3, c = 2, and d = - 19.

Then, the distance between the planes D = |ax₀+ by₀ + cz₀ + d| / √(a² + b² + c²).

D = |3*3 + 3*2 + 2*-3 + (-19)| / √(3² + 3² + 2²)

= |9 + 6 - 6  - 19| / √(9 + 9 + 4)

= |- 10| / √22

= 10/√22

= 2.132.

The distance between two parallel planes is 2.132.

Answer 2

Contd....

Question 2 :

The line is x = (- 5, 5, - 5) + t(- 20, - 5, 5) and the plane is x - 2y + 2z = 0.

From the given data the line and the plane do not intersect, means that, the line is parallel to the plane.

So, take any point on the line : x = (- 5, 5, - 5) + t(- 20, - 5, 5) and find its perpendecular distance from the plane : x - 2y + 2z = 0.

Any point on the line say  (x₀, y₀, z₀) = (- 5, 5, - 5)

From the plane, we can determined that, a = 1, b = - 2, c = 2, and d = 0.

Then, the perpendecular distance D = |ax₀+ by₀ + cz₀ + d| / √(a² + b² + c²).

D = |1*-5 + -2*5 + 2*-5 + 0| / √(1² + (-2)² + 2²)

= |- 5 - 10 - 10| / √(1 + 4 + 4)

= |- 25| / √9

= 25/3

= 8.33.

The distance between the line and the plane is 8.33.