Find the vector equation of the line that passes through the point (2,-1,7) and is parallel to the line of intersection of the planes x + 2y - 3z = -6 and 3x - y + 2z = 4
To find the vector equation of the line, we need two things: a point on the line and the direction of the line.
First, let's find a point on the line. We are given that the line passes through the point (2, -1, 7), so this will be our point on the line.
Next, we need to find the direction of the line. The direction of the line is parallel to the line of intersection of the two planes x + 2y - 3z = -6 and 3x - y + 2z = 4. To find the direction, we need to find the normal vector of the plane of intersection.
To find the normal vector of a plane, we look at the coefficients of x, y, and z in the plane equation. The normal vector of the first plane, x + 2y - 3z = -6, is (1, 2, -3). The normal vector of the second plane, 3x - y + 2z = 4, is (3, -1, 2).
Now, to find the direction of the line of intersection, we take the cross product of the two normal vectors. The cross product of (1, 2, -3) and (3, -1, 2) gives us the direction vector of the line of intersection.
The cross product is calculated as follows:
(a1, a2, a3) x (b1, b2, b3) = (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1)
Using this formula, we can calculate the direction vector:
(1, 2, -3) x (3, -1, 2) = (2 * 2 - (-3) * (-1), (-3) * 3 - 2 * 2, 1 * (-1) - 2 * 3)
= (4 - 3, -9 - 4, -1 - 6)
= (1, -13, -7)
Now, we have a point on the line (2, -1, 7) and the direction vector of the line (1, -13, -7). We can use these values to form the vector equation of the line in the form of (x, y, z) = (x0, y0, z0) + t(a, b, c):
(x, y, z) = (2, -1, 7) + t(1, -13, -7)
Therefore, the vector equation of the line that passes through the point (2, -1, 7) and is parallel to the line of intersection of the given planes is (x, y, z) = (2, -1, 7) + t(1, -13, -7).