Find the point of intersection and the angle between x - y = 32 and 3x - 8y = 6.

Question

calculus Find the point of intersection and the angle between x - y = 32 and 3x - 8y = 6.

 

Answer 1

The point of intersection is (50,18) which is obtained by solving the set of linear equations...since you are taking calculus, I will assume you know how to do that!

 

Now graph the 2 equations.

The first equation intersects the x axis at an angle of arc tan 1 (call it ∠1) and the second equation intersects the x axis at an angle of arc tan (3/8) (call it ∠2).

 

From geometry you will see that the angle of intersection of the 2 lines is ∠1 - ∠2.

Answer 2

If x-y = 32, then y = x-32

 

If 3x-8y=6, then -8y = -3x+6

 

                            y = (3/8)x - (3/4)

 

The lines intersect when x-32 = (3/8)x - (3/4)

 

                                  8x - 256 = 3x - 6

 

                                           5x = 250       x = 50 and y = 50-32 = 18

 

The lines intersect at (50, 18).

 

Another point on the line y = x-32 is (0, -32).

 

    The vector u = <50-0, 18-(-32)> = <50,50> is parallel to the line y = x-32.

 

Another point on the line y = (3/8)x - (3/4) is (2,0).

 

    The vector v = <50-2, 18-0> = <48, 18> is parallel to y = (3/8)x - (3/4).

 

The angle, θ, between the lines is the same as the angle between u and v.

 

   cosθ = (u•v) / [(llull)(llvll)

 

       u•v = (50)(48) + (50)(18) = 3300

 

       llull = √[50²+50²] = √5000    llvll = √2628

 

So, cosθ = 3300 / [√5000√2628]

 

             = 0.9103665

 

           θ = Cos^-1(0.9103665) = 24.4°