Answer:
a) To show that the lines L₁ and L₂ lie in the same plane, we can demonstrate that both lines satisfy the equation of the given plane P: 2x - y + 3z = 15.
For Line L₁:
The parametric equations of L₁ are:
x = 1 + 4t
y = 3 + 3t
z = t
Substituting these values into the equation of the plane:
2(1 + 4t) - (3 + 3t) + 3t = 15
2 + 8t - 3 - 3t + 3t = 15
7t - 1 = 15
7t = 16
t = 16/7
Therefore, Line L₁ satisfies the equation of plane P.
For Line L₂:
The parametric equations of L₂ are:
x = 1 + 8t
y = 2 + 6t
z = 3 + 2t
Substituting these values into the equation of the plane:
2(1 + 8t) - (2 + 6t) + 3(3 + 2t) = 15
2 + 16t - 2 - 6t + 9 + 6t = 15
16t + 6t + 6t = 15 - 2 - 9
28t = 4
t = 4/28
t = 1/7
Therefore, Line L₂ satisfies the equation of plane P.
Since both Line L₁ and Line L₂ satisfy the equation of plane P, we can conclude that they lie in the same plane.
The general equation of the plane P is 2x - y + 3z = 15.
b) To find the distance between Line L₁ and the Y-axis, we can find the perpendicular distance from any point on Line L₁ to the Y-axis.
Consider the point P₁(1, 3, 0) on Line L₁. The Y-coordinate of this point is 3.
The distance between the Y-axis and point P₁ is the absolute value of the Y-coordinate, which is 3.
Therefore, the distance between Line L₁ and the Y-axis is 3 units.
c) To find the point B on plane P that is closest to the point A(1, 0, 7), we can find the perpendicular distance from point A to plane P.
The normal vector of plane P is (2, -1, 3) (coefficient of x, y, z in the plane's equation).
The vector from point A to any point (x, y, z) on the plane can be represented as (x - 1, y - 0, z - 7).
The dot product of the normal vector and the vector from point A to the plane is zero for the point on the plane closest to point A.
(2, -1, 3) · (x - 1, y - 0, z - 7) = 0
2(x - 1) - (y - 0) + 3(z - 7) = 0
2x - 2 - y + 3z - 21 = 0
2x - y + 3z = 23
Therefore, the point B on plane P that is closest to point A(1, 0, 7) lies on the plane with the equation 2x - y + 3z = 23.