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Home Vectors Planes in Space Examples Example 12: Intersection Lines Between Planes
 
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Example 12: Intersection Lines Between Planes

Find the line L of intersection of the planes

4x - 5y + z = 2, x + 2z = 0.

Step 1

To get a position vector of L, we find any point on both planes. Setting z = 0 and solving for x and y, we obtain the point S(0, -2/5, 0) on both planes. Thus S = - (2/5)j is a position vector of L.

Step 2

To get a direction vector D of L we need a vector perpendicular to the normal vectors of both planes. The normal vectors are

M = 4i - 5j + k, N = i + 2k.

We take

10_vectors-191.gif

Thus L is the line X = -(2/5)j + t(-10i - 7j + 5k).

Home Vectors Planes in Space Examples Example 12: Intersection Lines Between Planes

Last Update: 2006-11-15