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Chapter 12: Problem 80

Let \(L_{1}\) be the line through the points \((1,2,6)\) and \((2,4,8)\) Let \(L_{2} b\) e the line of intersection of the planes \(P_{1}\) and \(P_{2}\) where \(P_{1}\) is the plane \(x-y+2 z+1=0\) and \(P_{2}\) is the plane through the points \((3,2,-1),(0,0,1),\) and \((1,2,1) .\) Calculate the distance between \(L_{1}\) and \(L_{2} .\)

Short Answer

Expert verified
The distance between \(L_1\) and \(L_2\) is approximately 6.77 units.

Step by step solution

01

Find Direction of Line \(L_1\)

The direction vector of line \(L_1\) through points \((1, 2, 6)\) and \((2, 4, 8)\) is found by subtracting the coordinates of the points: \((2-1, 4-2, 8-6) = (1, 2, 2)\). So the direction vector \( \mathbf{v_1} = (1, 2, 2) \).
02

Determine Equation for Plane \(P_2\)

First, find two vectors on plane \(P_2\) using the points given, \((0,0,1)\) and \((1,2,1)\), by subtracting point \((3,2,-1)\). Vector 1: \((0-3, 0-2, 1+1) = (-3, -2, 2)\). Vector 2: \((1-3, 2-2, 1+1) = (-2, 0, 2)\). Find the normal to the plane by cross product of these vectors: \((-3, -2, 2) \times (-2, 0, 2)\) yielding \(\mathbf{n_2} = (-4, -2, -4)\). Simplifying gives \(\mathbf{n_2} = (2, 1, 2)\). The plane equation is \(2x + y + 2z + D = 0\). Substitute the point \((3, 2, -1)\) to find \(D\). This gives \(2(3) + 2 + 2(-1) + D = 0\), so \(D = -6\). Thus, \(P_2\) is \(2x + y + 2z - 6 = 0\).
03

Find Direction Vector of Line \(L_2\)

The direction vector for line \(L_2\), the line of intersection, is the cross product of the normals of \(P_1\) and \(P_2\). Normal of \(P_1\) is \( (1, -1, 2) \), normal of \(P_2\) is \( (2, 1, 2) \). Cross product \((1, -1, 2) \times (2, 1, 2)\) results in \((-4, 2, 3)\). Simplifying gives the direction vector of \(L_2\) as \((-4, 2, 3)\).
04

Find and Use Point on Line \(L_2\)

Solve for a point on \(L_2\) by solving simultaneous equations \( x - y + 2z + 1 = 0 \) and \( 2x + y + 2z = 6 \). Adding these, \(3x + 4z = 5\). Let \(z = 0\); then \(x = \frac{5}{3}\) and substituting back gives \(y = \frac{-1}{3}\). Thus, a point on \(L_2\) is \(\left(\frac{5}{3},\frac{-1}{3},0\right)\).
05

Calculate Shortest Distance Between \(L_1\) and \(L_2\)

The vector from point on \(L_1\), say \((1, 2, 6)\), to \( \left(\frac{5}{3},\frac{-1}{3},0\right) \) on \(L_2\) is \( \left(\frac{5}{3} - 1, \frac{-1}{3} - 2, 0 - 6 \right) = \left(\frac{2}{3}, -\frac{7}{3}, -6\right)\). To find the shortest distance, calculate the projection of this vector on the cross product vector \((1, 2, 2) \times (-4,2,3)\), giving \((-2, 10, 10)\). Magnitude is \(|\mathbf{w}|^2 = 2^2 + 10^2 + 10^2 = 204\). Distance is \(\frac{|\left(\frac{2}{3}, -\frac{7}{3}, -6\right) \cdot (-2, 10, 10)|}{\sqrt{204}} = \frac{96}{\sqrt{204}} = 6.77\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Direction Vector
To understand the direction vector, imagine it as a guide that tells us which way a line is heading.
We can find a direction vector by taking any two points on the line and subtracting their corresponding coordinates.
  • For a line going through points \(A(1, 2, 6)\) and \(B(2, 4, 8)\), you find the direction vector \(\mathbf{v_1} = (2-1, 4-2, 8-6) = (1, 2, 2)\).
This vector \(\mathbf{v_1} = (1, 2, 2)\) represents the direction in which the line \(L_1\) is extending.
Having this vector is like having directions on a map that help us understand the orientation of \(L_1\).
Plane Intersection
The intersection of two planes creates a line, and finding this line can help solve many geometric puzzles.
To identify the line of intersection, calculate the normal vectors for both planes first.
A normal vector is perpendicular to the plane and provides a crucial clue in finding the intersection.
  • For plane \(P_1: x-y+2z+1=0\), its normal vector is \(\mathbf{n_1} = (1, -1, 2)\).
  • For plane \(P_2\), which passes through three points \((3,2,-1),(0,0,1), (1,2,1)\), calculate two vectors on the plane and take their cross product to find the normal \(\mathbf{n_2} = (2, 1, 2)\).
By crossing these normal vectors, you identify the direction of the line of intersection: \(\mathbf{v_2} = (-4, 2, 3)\).
This gives a clear picture of where and how the planes intersect in space.
Cross Product
The cross product is a special operation used with vectors that results in another vector perpendicular to the original pair.
This is extremely useful in finding lines of intersection between planes.
  • For vectors \(\mathbf{a} =(-3, -2, 2)\) and \(\mathbf{b} = (-2, 0, 2)\) on plane \(P_2\), their cross product \(\mathbf{a} \times \mathbf{b} =(2, 1, 2)\) reveals a vector perpendicular to both.
  • Another example is crossing normals \(\mathbf{n_1}=(1, -1, 2)\) and \(\mathbf{n_2}=(2, 1, 2)\), yielding the direction vector \(\mathbf{v_2}=(-4, 2, 3)\) for line \(L_2\).
The cross product is computed as\[(a_2b_3-a_3b_2, a_3b_1-a_1b_3, a_1b_2-a_2b_1)\]where \(\mathbf{a}=(a_1, a_2, a_3)\) and \(\mathbf{b}=(b_1, b_2, b_3)\).
It helps us establish relations among different spatial directions.
Vector Projection
Vector projection helps to measure how much one vector points in the direction of another.
It is essential in calculations involving distance, especially between skew lines.
  • Consider a vector \((\frac{2}{3}, -\frac{7}{3}, -6)\) indicating a path from a point on \(L_1\) to a point on \(L_2\).
  • Project this on the cross product vector \((-2, 10, 10)\) to obtain the shortest distance.
The projection provides
\[\text{proj}_{\mathbf{w}}\mathbf{v} = \frac{\mathbf{v}\cdot\mathbf{w}}{\mathbf{w}\cdot\mathbf{w}}\mathbf{w}\]where \(\mathbf{v} = (\frac{2}{3}, -\frac{7}{3}, -6)\) and \(\mathbf{w}=(-2, 10, 10)\).
Computing this ensures the most accurate calculation of distance, providing the solution's clarity and usefulness.

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Most popular questions from this chapter

Determine whether the lines \(L_{1}\) and \(L_{2}\) are parallel, skew, or intersecting. If they intersect, find the point of intersection. $$ \begin{array}{l}{L_{1}: x=5-12 t, \quad y=3+9 t, \quad z=1-3 t} \\ {L_{2}: x=3+8 s, \quad y=-6 s, \quad z=7+2 s}\end{array} $$

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