How do you find the equation of a plane that passes through the point (0, 7, 7) and contains the line represented by the parametric equations x = 1 + 3t, y = 2 + 32t, z = 21 + t?

by

To find the equation of a plane that passes through a specific point and contains a line, we’ll start by determining key components of the problem. We know the plane should pass through the point (0, 7, 7) and it contains the line given by the parametric equations:

  • x = 1 + 3t
  • y = 2 + 32t
  • z = 21 + t

First, let’s extract some information from the line. The parametric equations describe a line in three-dimensional space. We can identify two points that lie on this line by choosing specific values for the parameter t.

1. For t = 0:

x = 1, y = 2, z = 21
ightarrow (1, 2, 21)

2. For t = 1:

x = 1 + 3(1) = 4, y = 2 + 32(1) = 34, z = 21 + 1 = 22
ightarrow (4, 34, 22)

Now, we have three points to work with:

  • Point A: (0, 7, 7) – the point through which the plane passes
  • Point B: (1, 2, 21) – a point on the line
  • Point C: (4, 34, 22) – another point on the line

Next, we find the vectors that lie in the plane:

  • Vector AB: From A to B

    AB = B – A = (1 – 0, 2 – 7, 21 – 7) = (1, -5, 14)
  • Vector AC: From A to C

    AC = C – A = (4 – 0, 34 – 7, 22 – 7) = (4, 27, 15)

To find the normal vector of the plane, we take the cross product of vectors AB and AC:

Let AB = (1, -5, 14) and AC = (4, 27, 15). Using the determinant method:

|i   j   k|
|1  -5  14|
|4  27  15|

Calculating the components:

i: (-5 * 15 - 14 * 27) = -75 - 378 = -453
j: -(1 * 15 - 14 * 4) = -(15 - 56) = 41
k: (1 * 27 - (-5) * 4) = 27 + 20 = 47

Thus, the normal vector N of the plane is (-453, 41, 47).

Now, we can use the point-normal form of the equation of a plane, which is:

A(x - x_0) + B(y - y_0) + C(z - z_0) = 0

Here, (x_0, y_0, z_0) is the point through which the plane passes, and (A, B, C) are the components of the normal vector. Plugging in our values:

-453(x - 0) + 41(y - 7) + 47(z - 7) = 0

Expanding this equation gives:

-453x + 41y - 287 + 47z - 329 = 0

So, simplifying the plane equation results in:

-453x + 41y + 47z - 616 = 0

This is the equation of the plane that passes through the point (0, 7, 7) and contains the given line.

Leave a Comment