Find the vector equation of the line of intersection of the planes 2x+y-z=4 and 3x+5y+2z=13.

The line of intersection will be parallel to both planes. We can write the equations of the two planes in 'normal form' as r.(2,1,-1)=4 and r.(3,5,2)=13 respectively. We can then read off the normal vectors of the planes as (2,1,-1) and (3,5,2). The vector product of these two normals will give a vector which is perpendicular to both normals and hence parallel to both planes. Finding the vector product: (2,1,-1)^(3,5,2) = (1x2-5x-1, -1x3-2x2, 2x5-1x3) = (7,-7,7) =7(1,-1,1). The vector equation of a line is r=a+kb where a is an arbitrary point on the line, k is a scalar and b is the direction vector of the line. To find a point on the line we could substitute z=0 (for example, any value would do) into the equations of the planes and solve the resulting simultaneous equations for x and y (the line will lie in both planes). This gives the equations 2x+y=4 and 3x+5y=13 which solve to give x=1 and y=2. The equation for the line of intersection is therefore r=(1,2,0)+k(1,-1,1).