Question

Find an equation of the plane with the given characteristics. The plane contains the lines given by x - 4 -3 = y - 4 = z and x - 2 -1 = y - 1 4 = z - 2 -1?

Answer 1

To find the equation of the plane containing the given lines, we need to determine two points on each line and find the vector equation of the plane. Using the points A, B, and C, we can write the vector equation as r = a + sv + tw.

Step-by-step explanation:

To find an equation of the plane containing the given lines, we need to determine two points on each line and find the vector equation of the plane. Let's find the two points for each line first:

Line 1: x - 4 = y - 4 = z, point A(4, 4, 0) and point B(0, 0, 0).

Line 2: x - 2 = y - 1 = z - 2, point C(2, 1, 2) and point D(0, 1, 0).

Now, we can find the vector equation of the plane using the points A, B, and C:

Let AB = v = B - A = (-4, -4, 0). Let AC = w = C - A = (-2, -3, 2).

The vector equation of the plane is r = a + sv + tw, where a is a point on the plane and s, t are scalar parameters.

Substituting the values, we get r = (4, 4, 0) + s(-4, -4, 0) + t(-2, -3, 2), which simplifies to:

r = (4 - 4s - 2t, 4 - 4s - 3t, 2t).