Jennifer Y. answered • 11/23/19
Math Teacher in High School
You need to find "direction" vector (d) and a point vector (a) passing through the intersection line.
Then, the line equation can be a + td (t∈R)
1) Direction vector is the cross product of two normal vector of each plane.
Normal vector of x+y+z=1 is (1,1,1) and normal vector of x-y-z=2 is (1,-1,-1)
Direction vector is (-1-(-1), -(-1-1), -1-1) = (0, 2, -2)
2) A point vector passing through both planes should be on the intersection line. There are numerous points, so randomly pick one by letting z=0.
Let z=0 and solve x+y=1 and x-y=2. Then, x=3/2, y=-1/2
∴ The point vector is (3/2, -1/2, 0)
By 1) and 2), the line equation is (3/2, -1/2, 0) + t (0, 2, -2)
Jennifer Y.
Because a normal vector to a plane is perpendicular to the plane. Since we are finding for the line that lies on both planes, the direction vector should be perpendicular to both normal vectors, which is the definition of "cross product." "Cross product" vector is defined as a vector c that is perpendicular (orthogonal) to both a and b. If you search for the image of cross product, it might help you understand what it is. You're welcome!11/23/19
Ashley P.
How do we get the normal.vector of a plane equation of the type x+y+z=2?11/23/19
Jennifer Y.
Always the coefficient of x, y, z should be the normal vector.11/23/19
Ashley P.
Even if when theres a value on the other of the equation mark??11/23/19
Jennifer Y.
The value on the right hand side is not related with the normal vector, rather it relates the point on the plane. In general, the plane equation with normal vector (a,b,c) and a point (x',y',z') on it is a(x-x')+b(y-y')+c(z-z')=0. If you expand, you can have a value on the right hand side and the value depends on (x',y',z'). Imagine that there are numerous planes that are orthogonal to the normal vector in a space. If a specific point on the plane is given, now you can determine a specific plane out of the numerous planes. That's why you need both normal vector and a point is needed for a plane equation. It is like slope-point form in 2-dim.11/23/19
Ashley P.
Thanks a lot for clearing the confusion11/23/19
Ashley P.
Thank you for the explanation! Why do we take the cross product as the direction vector?11/23/19