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11.5 Lines and planes
When two points are defined in space there are an infinite number of vectors through those two points. To define the unique line through the points we need to find all those ordered triples that define a vector through those points. For example, if the points are ( 1, 1, 2 ) and ( 2, 4, 8 ), then the vector v = < 1, 3, 6 > is parallel to the line through the points. All vectors parallel to v, the direction vector, are parallel to the line. The line is defined parametrically by the set of all ( x, y, z ) such that x = 1 + 1t, y = 1 + 3t and z = 2 + 6t. Please note that these are all those ordered triples that define parallel vectors and contain the point ( 1, 1, 2 ). We could solve for the parameter t and set ( x - 1 )/1 = ( y - 1 )/3 = ( z - 2 )/6, the symmetric form of the line, but generally we will use the parametric form.
To define a plane through 3 noncollinear points, P, Q and R, we will first find a vector, v = < a, b, c >, orthogonal to the two vectors PQ and PR. Of course this is their cross product. Then we will find all those ordered triples X = ( x, y, z ) such that the vector PX is orthogonal to v, that is, their dot product is 0. ( We could have used point Q or point R instead of point P. ) If the coordinates of point P are ( x1, y1, z1 ) we have PX = < x - x1, y - y1, z - z1 >. Thus the dot product of PX and v is then
a( x - x1 ) + b( y - y1 ) + c( z - z1 ) = 0. This is the equation of the plane containing point ( x1, y1, z1 ) normal ( perpendicular ) to the vector < a, b, c >.
Let's do an example. Given the 3 points P( 3, 0, 0 ), Q( 0, 1, 0 ) and R( 0, 0, 2 ), the equation of lines PQ, PR and QR are as follows. PQ: x =3 - 3t, y = 0 + t, z = 0 + 0t; PR: x = 3 - 3t, y = 0 + 0t, z = 0 + 2t; and QR: x =0 + 0t, y = 1 - t, z = 0 + 2t. The cross product of PQ and PR is < 2, 6, 3 >. This is normal to the plane. The plane has an equation of 2( x - 3 ) + 6( y - 0 ) + 3( z - 0 ) = 0.Here are the graphs.
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When two planes intersect their intersection is a line. The angle of intersection is the same angle formed by their normal vectors, the cosine of which is their dot product divided by the product of their magnitudes. To find the equation of the line of intersection calculate the cross product of their normals. This vector orthogonal to the normals will be parallel to the intersection line. One point on the line of intersection can be found by setting x ( or y or z ) equal to 0.
For example, given the planes 2x + 6y + 3z = 6 and 3x - 3y + 4z = 0 let's find the cosine of the angle of intersection and the equation of the line of intersection. The two normals are < 2, 6, 3 > and < 3, -3, 4 >, resp. Thus the cosine of the angle between the planes is 0 and the planes are perpendicular.
Crossing the two normals gives < 33, 1, -24 >. This vector is parallel to the line of intersection. To find a point on the line of intersection, set x = 0 and solve 6y + 3z = 6 and -3y + 4z = 0 simultaneously. This gives z = 6/11 and y = 8/11. The equation of the intersecion line in parametric form is: x = 0 + 33t, y = 8/11 + t and z = 6/11 - 24t. Every point on this line belongs to both planes. By letting t = 3/11 we obtain the point ( 9, 1, -6 ) and could use the line: x = 9 + 33t, y = 1 + t and z = -6 - 24t.
