Math Problem Statement

Find a plane containing the line r(t) = <2, -4, -3> + t<-4, -1, 5> and orthogonal to the plane 8x + 1y - 7z = 8.

Solution

Let's analyze the given problem:

Problem statement:

You need to find a plane that contains the line:

r(t)=2,4,3+t4,1,5\mathbf{r}(t) = \langle 2, -4, -3 \rangle + t \langle -4, -1, 5 \rangle

and is orthogonal to the plane given by the equation:

8x+1y7z=88x + 1y - 7z = 8

Step-by-step solution:

1. Find the direction vector of the line:

The vector associated with the line r(t)\mathbf{r}(t) is given by the direction vector 4,1,5\langle -4, -1, 5 \rangle.

2. Normal vector to the given plane:

The normal vector to the plane 8x+1y7z=88x + 1y - 7z = 8 is n=8,1,7\mathbf{n} = \langle 8, 1, -7 \rangle.

3. Normal vector to the desired plane:

The plane we need to find is orthogonal to the given plane. This means the normal vector of the desired plane must be perpendicular to the normal vector of the given plane.

Let the normal vector to the desired plane be N=a,b,c\mathbf{N} = \langle a, b, c \rangle. Since this plane is orthogonal to the plane 8x+y7z=88x + y - 7z = 8, their normal vectors must be perpendicular. Thus, the dot product of n=8,1,7\mathbf{n} = \langle 8, 1, -7 \rangle and N=a,b,c\mathbf{N} = \langle a, b, c \rangle must be zero:

8a+1b7c=08a + 1b - 7c = 0

4. Plane containing the line:

The plane also needs to contain the line. Since the direction vector of the line is 4,1,5\langle -4, -1, 5 \rangle, this vector must lie on the plane. The direction vector of the line and the normal vector of the plane must also be perpendicular. So, the dot product of the normal vector a,b,c\langle a, b, c \rangle and the line’s direction vector 4,1,5\langle -4, -1, 5 \rangle should also be zero:

4a1b+5c=0-4a - 1b + 5c = 0

5. Solving the system of equations:

We now have a system of two equations:

8a+b7c=08a + b - 7c = 0 4ab+5c=0-4a - b + 5c = 0

Solve this system to find aa, bb, and cc.

6. The point on the line:

The point 2,4,3\langle 2, -4, -3 \rangle lies on both the line and the plane. Thus, the equation of the plane can be written as:

a(x2)+b(y+4)+c(z+3)=0a(x - 2) + b(y + 4) + c(z + 3) = 0

After solving the system, you can substitute aa, bb, and cc into this equation to get the final plane equation.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Planes
Lines in 3D
Dot Product
Normal Vectors

Formulas

Vector equation of the line: r(t) = r0 + t * v
Equation of a plane: a(x - x0) + b(y - y0) + c(z - z0) = 0
Dot product condition for perpendicularity: n1 ⋅ n2 = 0

Theorems

Orthogonality of vectors
Dot product theorem
Plane equation from normal vector

Suitable Grade Level

Grades 11-12, College Level