Find parametric equations for the line
through the point P(2, 9, 4) that is parallel to the plane x + y + z = 6 and perpendicular to the line x = 5 + t , y = 5 − t , z = 5t .
Answer Choices:
1. x = 2 + 6t, y = 9 + 4t, z = 4 + t
2. x = 2 + t, y = a + t, z = 4 − 2t
3. x = 2 − 6t, y = 9 + 4t, z = 4 + t
4. x = 2 − 6t, y = 9 + 4t, z = 4 + 2t
5. x = 2 + 6t, y = 9 + 4t, z = 4 + 2t
To find the parametric equations for the line, we need to consider the given conditions.
First, the line should pass through the point P(2, 9, 4). This means that any point on the line can be represented as (x, y, z) = (2, 9, 4)
Secondly, the line should be parallel to the plane x + y + z = 6. This means that the direction vector of the line should be perpendicular to the normal vector of the plane.
The normal vector of the plane x + y + z = 6 can be obtained by taking the coefficients of x, y, and z. It is (1, 1, 1).
Next, we consider that the line should be perpendicular to the given line x = 5 + t, y = 5 − t, z = 5t. Hence, the direction vector of the line should be perpendicular to the direction vector of this given line.
The direction vector of the given line is (1, -1, 5).
To find a direction vector that is both perpendicular to (1, 1, 1) and (1, -1, 5), we can take the cross product of these two vectors.
Cross product: (1, 1, 1) × (1, -1, 5)
= (6, -4, -2)
Therefore, the direction vector of the line is (6, -4, -2).
To get the parametric equations, we can express the coordinates (x, y, z) in terms of a parameter t.
x = 2 + 6t
y = 9 - 4t
z = 4 - 2t
Comparing these equations to the given answer choices, we can see that the correct choice is:
5. x = 2 + 6t, y = 9 + 4t, z = 4 + 2t