# Vectors

• Sep 8th 2009, 04:37 AM
tdat1979
Vectors
(a) Find parametric equations for the line of intersection between the planes x + y = 0 and 3y + 8z = 6.

(b) Find the point of intersection of the line found in part (a) and the plane 2x + 5y + 4z = 5.
• Sep 8th 2009, 04:50 AM
galactus
a: $n_{1}=(1,1,0), \;\ n_{2}=(3,0,8)$

These are the normals to the planes.

Find the cross product: $n_{1}\times n_{2}=(8,-8,-3)$

Now, let z=0 in the two equations and solve for x and y.

Finish?.
• Sep 8th 2009, 05:20 AM
HallsofIvy
Another way: the two planes are x + y = 0 and 3y + 8z = 6. Solve for, say, x and z in terms of y: x= -y and 8z= 6- 3y so z= (3/4)- (3/8)y.

Now take t= y/8 as parameter. x= -8t, y= 8t, z= (3/4)- 3t.

To find the point where that line intersects the plane 2x + 5y + 4z = 5, replace x, y, and z in that equation by their parametric expressions: 2(-8t)+ 5(8t)+ 4((3/4)- 3t)= -16t+ 40t- 12t+ 3= 12t+ 3= 5. 12t= 2, t= 1/6. Now put that value of t back into the parametric equations: x= -8(1/6)= -4/3, y= 8(1/6)= 4/3, z= (3/4)- 3(1/6)= 3/4- 2/4= 1/4. The point of intersection is (-4/3, 4/3, 1/4).

Note that those values of x, y, and z satisfy the equations of all three planes. -4/3+ 4/3= 0, 3(4/3)+ 8(1/4)= 4- 2= 6, and 2(-4/3)+ 5(4/3)+ 4(1/4)= -8/3+ 20/3+ 1= 12/3+ 1= 4+ 1= 5.
• Sep 8th 2009, 04:18 PM
tdat1979
I'm confused.