Math Problem Statement

A line L1 has equation r = (-5, -3, 2) + λ(-1, 2, 2). A line L2 passing through the origin intersects L1 and is perpendicular to L1. (a) Find a vector equation of L2.

Solution

To solve this problem, let’s go through the given information and find the vector equation of line L2L_2 as required.

Given Data

  1. Equation of L1L_1: r=(532)+λ(122)\mathbf{r} = \begin{pmatrix} -5 \\ -3 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 2 \\ 2 \end{pmatrix} where λ\lambda is a parameter. Here, the direction vector of L1L_1 is: d1=(122).\mathbf{d_1} = \begin{pmatrix} -1 \\ 2 \\ 2 \end{pmatrix}.

  2. Conditions for L2L_2:

    • L2L_2 passes through the origin (0, 0, 0).
    • L2L_2 is perpendicular to L1L_1.
    • L2L_2 intersects L1L_1.

Solution

To find the vector equation of L2L_2, we need a direction vector d2\mathbf{d_2} for L2L_2 that is perpendicular to d1\mathbf{d_1}.

Since L2L_2 is perpendicular to L1L_1, the direction vector d2\mathbf{d_2} of L2L_2 should satisfy: d1d2=0\mathbf{d_1} \cdot \mathbf{d_2} = 0 This gives us the perpendicularity condition.

  1. Direction Vector for L2L_2:
    Let d2=(xyz)\mathbf{d_2} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}. Then: (122)(xyz)=0\begin{pmatrix} -1 \\ 2 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} = 0 Expanding this dot product, we get: x+2y+2z=0-x + 2y + 2z = 0 This equation represents a constraint on the values of xx, yy, and zz for d2\mathbf{d_2}.

  2. Choosing Values for d2\mathbf{d_2}:
    To satisfy this equation, we can select values for xx, yy, and zz that solve x+2y+2z=0-x + 2y + 2z = 0. One possible choice is: d2=(211)\mathbf{d_2} = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} since: 2+21+2(1)=0-2 + 2 \cdot 1 + 2 \cdot (-1) = 0

  3. Equation of L2L_2: Since L2L_2 passes through the origin, its position vector is: r=μ(211)\mathbf{r} = \mu \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} where μ\mu is a parameter.

Thus, the vector equation of L2L_2 is: r=μ(211)\mathbf{r} = \mu \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}

Final Answer

The vector equation of L2L_2 is: r=μ(211),μR\mathbf{r} = \mu \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}, \quad \mu \in \mathbb{R}

Would you like more details or have any questions about this solution?


Here are some related questions you might find interesting:

  1. How do we confirm if two lines in three-dimensional space intersect?
  2. How can we determine the angle between two lines in three-dimensional space?
  3. What is the general method for finding a line perpendicular to a given line in 3D space?
  4. How do we find the shortest distance between two skew lines?
  5. How does the dot product help in determining perpendicularity in vector equations?

Tip: When working with perpendicular vectors, using the dot product is a powerful tool to ensure orthogonality!

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Math Problem Analysis

Mathematical Concepts

Vectors
Lines in 3D space
Dot Product
Perpendicularity

Formulas

Vector equation of a line: r = a + λd
Perpendicularity condition: d1 · d2 = 0

Theorems

Perpendicularity theorem in 3D space

Suitable Grade Level

Grades 11-12