Chapter 11: Problem 56
Let \(L_{1}\) and \(L_{2}\) be the lines whose parametric equations are $$ \begin{array}{lll} L_{1}: x=4 t, & y=1-2 t, & z=2+2 t \\ L_{2}: x=1+t, & y=1-t, & z=-1+4 t \end{array} $$ (a) Show that \(L_{1}\) and \(L_{2}\) intersect at the point \((2,0,3)\). (b) Find, to the nearest degree, the acute angle between \(L_{1}\) and \(L_{2}\) at their intersection. (c) Find parametric equations for the line that is perpendicular to \(L_{1}\) and \(L_{2}\) and passes through their point of intersection.
Short Answer
Step by step solution
Determine Intersection Point for L1
Solve for t in L1
Determine Intersection Point for L2
Solve for t in L2
Find Direction Vectors
Compute Dot Product
Magnitude of Vectors
Find Cosine of Angle
Calculate Angle
Find Perpendicular Direction Vector
Write Parametric Equations
Simplifying Cross Product
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Equations
- For \(L_1\), the equations are \(x = 4t\), \(y = 1 - 2t\), and \(z = 2 + 2t\).
- For \(L_2\), the equations are \(x = 1 + t\), \(y = 1 - t\), and \(z = -1 + 4t\).
This approach allows for complex shapes and systems to be broken down into simpler linear components, which are easier to analyze and compute.
Intersection of Lines
- Start by setting the parametric equations for the two lines equal to each other for each coordinate axis.
- The aim is to find a common \(t\) (or distinct parameters for each line) that satisfies all the equations.
- For \(L_1\), \(t = \frac{1}{2}\)
- For \(L_2\), \(t = 1\)
Angle Between Vectors
- First, derive direction vectors from the parametric equations.
- Calculate the dot product, \(\mathbf{a} \cdot \mathbf{b}\), which combines these vectors in a scalar manner reflecting their directional relationship.
- Compute the magnitude of each vector to understand their specific lengths.
- Finally, use these to find the cosine of the angle \( \theta \) using the formula: \(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \).
Cross Product
- The cross product \(\mathbf{a} \times \mathbf{b}\) results in a new vector \(\mathbf{n}\) that is orthogonal to both \(\mathbf{a}\) and \(\mathbf{b}\).
- This vector provides a basis for defining a line perpendicular to both \(L_1\) and \(L_2\). \(\mathbf{n}\) becomes \(\langle -10, -14, 2 \rangle\) from the computation.
Ultimately, this process enables you to solve complex geometrical problems involving multiple dimensions and orientations, showcasing the cross product's power in vector calculus.