To find the vector equation of the plane containing the two intersecting lines, we can first find the normal vector of the plane by taking the cross product of the direction vectors of the two lines. The normal vector will be orthogonal to both direction vectors and thus will be parallel to the plane.
Direction vector of the first line: (2, 4, 3)
Direction vector of the second line: (-1, -1, 3)
Taking the cross product of these two vectors, we get:
(2, 4, 3) x (-1, -1, 3) = (9, -3, -6)
This vector is orthogonal to both direction vectors and thus is parallel to the plane. To find the vector equation of the plane, we can use the point-normal form of the equation, which is:
N · (r - P) = 0
where N is the normal vector, r is a point on the plane, and P is a known point on the plane. We can choose either of the two given points on the intersecting lines as the point P.
Let's use the point (4, 7, 3) on the first line as the point P. Then the vector equation of the plane is:
(9, -3, -6) · (r - (4, 7, 3)) = 0
Expanding and simplifying, we get:
9(x - 4) - 3(y - 7) - 6(z - 3) = 0
Simplifying further, we get:
9x - 3y - 6z = 0
Dividing by 3, we get:
3x - y - 2z = 0
Therefore, the vector equation of the plane containing the two intersecting lines is:
(3, -1, -2) · (r - (4, 7, 3)) = 0
or equivalently,
3x - y - 2z = 0.