A plane is perpendicular to both [x,y,z]=[1,-1,8]+s[1,2,-1] and [x,y,z]=[2,1,-3] and contains the point P(-1,4,2). Determine if the point A(7,20,16) is also on the plan

To determine if the point A(7, 20, 16) lies on the plane defined by the given conditions, we need to first establish the equation of the plane. The plane is perpendicular to a line defined by the vector equation and also contains a specific point. Let's break this down step by step.

Identifying the Direction Vector

The first line is given by the equation:

[x, y, z] = [1, -1, 8] + s[1, 2, -1]

From this, we can extract the direction vector of the line, which is:

• Direction vector: **d1 = [1, 2, -1]**

Finding the Normal Vector

The plane is perpendicular to this line, meaning the normal vector of the plane, **n**, can be taken as the same as the direction vector of the line:

• Normal vector: **n = [1, 2, -1]**

Using the Point on the Plane

The plane also contains the point P(-1, 4, 2). We can use the point-normal form of the plane equation, which is given by:

(x - x0, y - y0, z - z0) · n = 0

Here, (x0, y0, z0) is the point P(-1, 4, 2) and n is the normal vector [1, 2, -1]. Plugging in these values, we get:

(x + 1, y - 4, z - 2) · [1, 2, -1] = 0

Expanding this, we have:

• (x + 1) * 1 + (y - 4) * 2 + (z - 2) * (-1) = 0

Which simplifies to:

x + 1 + 2y - 8 - z + 2 = 0

Combining like terms gives us:

x + 2y - z - 5 = 0

Checking Point A(7, 20, 16)

Now, we need to see if the point A(7, 20, 16) satisfies this plane equation. We substitute x = 7, y = 20, and z = 16 into the equation:

7 + 2(20) - 16 - 5 = 0

Calculating this step-by-step:

• 7 + 40 - 16 - 5 = 0
• 31 ≠ 0

Conclusion

Since the left side does not equal zero, point A(7, 20, 16) does not lie on the plane defined by the given conditions. Therefore, A is not on the plane.