Determine a direction vector of the line of intersection of the two planes 2x - 7y + z = 5 and x + 3y - 6z = 2. If there is not a single line of intersection, classify the system.
since the planes have different direction numbers, they must intersect in a line.
The cross product of the normals to the planes will provide the line of intersection.
To determine a direction vector of the line of intersection of the two planes, we can find the normal vectors of both planes and take their cross product.
First, let's rewrite the equations of the planes in the general form Ax + By + Cz = D:
Plane 1: 2x - 7y + z = 5 --> 2x - 7y + z - 5 = 0 --> A1 = 2, B1 = -7, C1 = 1, D1 = -5
Plane 2: x + 3y - 6z = 2 --> x + 3y - 6z - 2 = 0 --> A2 = 1, B2 = 3, C2 = -6, D2 = -2
Now, the normal vector of each plane is given by the coefficients (A, B, C). Therefore, the normal vectors are:
Plane 1: n1 = (2, -7, 1)
Plane 2: n2 = (1, 3, -6)
To find the direction vector of the line of intersection, we take the cross product of the two normal vectors:
n = n1 x n2
Using the cross product formula:
n = [(B1 * C2) - (B2 * C1), (C1 * A2) - (C2 * A1), (A1 * B2) - (A2 * B1)]
n = [(-7 * (-6)) - (3 * 1), (1 * (-6)) - (-7 * 1), (2 * 3) - (1 * (-7))]
n = [33, -1, 17]
Therefore, the direction vector of the line of intersection of the two planes is (33, -1, 17).
In this case, there is a single line of intersection between the two planes, so we do not need to classify the system.