# Find the value of a and b (linear algebra)?

1. Apr 8, 2012

### BlueRope

1. The problem statement, all variables and given/known data
equation of line 1: (x-2)/2 = (y+6)/3 = (z-a)/-1

equation of line 2: (x+4)/5 = (y-6)/b = z-3

both lines are perpendicular to each other AND are in the same plane (coplanar)

ALSO find the equation of the plane

2. Relevant equations

3. The attempt at a solution

b = -3 because we know that the dot product of both lines = 0, but can't find the value of a

i tried resolving the system of equation of both lines using their parametric equations, but it doesn't work.

2. Apr 8, 2012

### HallsofIvy

Staff Emeritus
Okay, a vector in the direction of the first line is <2, 3, -1> and a vector in the direction of the second line is <5, b, 1> so they will be perpendicular if and only if the dot product, 10+ 3b- 1= 0. Yes,that gives 3b= -9, b= -3, as you say.

Now, two lines, that happen to be in the same plane, are either parallel or intersect. Since these lines are perpendicular, they cannot be parallel and must intersect. We can write the equations of the lines in parametric equations by taking each of the fractions for the first line to be equal to "t", and for the second to be equal to "s":
(x- 2)/2= t so x= 2t+ 2, (y+6)/3= t so y= 3t- 6, and (z- a)/-1= t so z= a -t.

(x+ 4)/5= s so x= 5s- 4, (y- 6)/b= (y- 6/(-3)= s so y= 6- 3s, and z- 3= s so z= s+ 3.

Solve the x and y equations for s and t, then put them into the z equations. For what value of a do they give the same z- value? What are x, y, z for the point of intersection?

Finally, you know two vectors in the plane so you know their cross product is perpendicular to the plane. And you know one point in the plane, the point of intersection of the two planes.