Analytic Geometry Revisited...

Question

You are given the planes (Π₁) , (Π₂) and the line (L₁)

(Π₁) : 2x + 3y +z = 6

(Π₂) : 3x + 5y +2z = 10

(L₁) : x =4 +t , y = 4 - 7t , z = 1 - 8t.

1. Find the equation of the plane (Π₃) that contains the line (L₁) orthogonal to both (Π₁) and , (Π₂)
2. Find the area of the triangle the three planes form, with one side being a segment of line (L₁).
3. Find the angle of intersection of the planes (Π₁) and , (Π₂).

Answer 1

i j k

2 3 1

3 5 2

i - j + k

The plane is x - y + z = ?

Substitute the coordinate (4 +t, 4 - 7t, 1 - 8t) and we get 4+t - (4 - 7t) + (1 - 8t) = 4 + t - 4 + 7t + 1 - 8t = 1

The lack of t tells us that the line is on the plane.

Then, (4 + t, 4 - 7t, 1 - 8t) meets 2x + 3y + z = 6 at t = ?

Substitute: 2(4 + t) + 3(4 - 7t) + (1 - 8t) = 6

8 + 2t + 12 - 21t + 1 - 8t = 6

21 - 27t = 6

15 = 27t

t = 5/9

The intersection is at (4 5/9, 1/9, -3 4/9)

(4 + t, 4 - 7t, 1 - 8t) meets 3x + 5y + 2z = 10 at t = ?

3(4+t) +5(4-7t)+2(1-8t) = 12 + 3t + 20 -35t + 2 - 16t = 34 - 48t

34 - 48t = 10

-48t = -24

t = 1/2

The intersection is at (4 1/2, 1/2, -3)

The 3 lines meet at

2x + 3y + z = 6

3x + 5y + 2z = 10

1x - 1y z = 1

which has solution

(1, 1, 1)

A = (4 5/9, 1/9, -3 4/9)

B = (4 1/2, 1/2, -3)

C = (1, 1, 1)

B - A = (-1/18, 7/18, 4/9)

C - A = (-3 5/9, 8/9, 4 4/9)

B-A x C-A =

i j k

-1/18 7/18 4/9

-3 5/9 8/9 4 4/9

has determinant

4/3 i - 4/3 j + 4/3 k

Thus, 1/2 ||B-A x C-A|| =

1/2(4/3 √3) =

2√3/3

We calculate the angle of intersection as we calculate the angle of intersection of lines, using the inner product

cos^-1 (2,3,1).(3,5,2)/(||(2, 3, 1)||*||(3,5,2)||) =

cos^-1 23/√(14*38) = 4.31º = 4º18'