Question

The parametric equations for a line L ₁ are as follows:

x=−3−3t
y=4−9t
z=−3+3t
Let L₂ be the line parallel to L ₁ and passing through the point (−2,4,2). Find the point P on L₂ whose x-coordinate is -1 P=(−1,0,0)

Answer 1

To find the point P on line L₂ parallel to L₁, passing through a point and with an x-coordinate of -1, we use the direction vector of L₁. By solving the parametric equations for the given x-coordinate, we determine that point P is (-1, 7, 1).

Step-by-step explanation:

The question asks us to find the point P on a line L₂, which is parallel to line L₁, given by parametric equations, and which passes through the point (-2,4,2). The provided point P should have an x-coordinate of -1. Since L₂ is parallel to L₁, it will have the same direction vector, which is given by the coefficients of t in L₁'s equations (i.e., -3, -9, 3).

To find the parametric equations for L₂, we use the direction vector of L₁ and the given point through which L₂ passes. The parametric equations for L₂ will therefore be:

• x = -2 - 3t
• y = 4 - 9t
• z = 2 + 3t

Since we want the x-coordinate of point P to be -1, we plug this value into the first equation to solve for t: -1 = -2 - 3t, which simplifies to -3t = 1, giving us t = -1/3. We then plug t = -1/3 into the other two equations to find the y-coordinate and z-coordinate of the point P. Doing so gives us y = 4 - 9(-1/3) = 7 and z = 2 + 3(-1/3) = 1. Hence, point P on L₂ is (-1, 7, 1).