Math Problem Statement

Let r1(t) = ⟨-2, 4, 7⟩ + t⟨0, 1, 3⟩ and r2(s) = ⟨-11, 2, 13⟩ + s⟨-3, 0, 4⟩. Find the point of intersection P of the two lines r1 and r2.

Solution

We are given two vector equations for the lines r1(t)\mathbf{r}_1(t) and r2(s)\mathbf{r}_2(s) as follows:

r1(t)=2,4,7+t0,1,3\mathbf{r}_1(t) = \langle -2, 4, 7 \rangle + t \langle 0, 1, 3 \rangle r2(s)=11,2,13+s3,0,4\mathbf{r}_2(s) = \langle -11, 2, 13 \rangle + s \langle -3, 0, 4 \rangle

To find the point of intersection PP of the two lines, we need to set the vector equations equal to each other because at the point of intersection, both lines share the same position vector. So:

r1(t)=r2(s)\mathbf{r}_1(t) = \mathbf{r}_2(s)

Expanding both equations:

r1(t)=2,4,7+t0,1,3=2,4+t,7+3t\mathbf{r}_1(t) = \langle -2, 4, 7 \rangle + t \langle 0, 1, 3 \rangle = \langle -2, 4 + t, 7 + 3t \rangle r2(s)=11,2,13+s3,0,4=113s,2,13+4s\mathbf{r}_2(s) = \langle -11, 2, 13 \rangle + s \langle -3, 0, 4 \rangle = \langle -11 - 3s, 2, 13 + 4s \rangle

Now, equate the corresponding components:

  1. 2=113s-2 = -11 - 3s
  2. 4+t=24 + t = 2
  3. 7+3t=13+4s7 + 3t = 13 + 4s

Solve these three equations for tt and ss.

Step 1: Solve the second equation for tt

4+t=2    t=24=24 + t = 2 \implies t = 2 - 4 = -2

Step 2: Solve the first equation for ss

2=113s    2+11=3s    9=3s    s=3-2 = -11 - 3s \implies -2 + 11 = -3s \implies 9 = -3s \implies s = -3

Step 3: Check consistency using the third equation

7+3t=13+4s7 + 3t = 13 + 4s Substitute t=2t = -2 and s=3s = -3 into the equation:

7+3(2)=13+4(3)    76=1312    1=17 + 3(-2) = 13 + 4(-3) \implies 7 - 6 = 13 - 12 \implies 1 = 1

Since the third equation is consistent, the solution is valid.

Step 4: Find the point of intersection PP

Substitute t=2t = -2 into r1(t)\mathbf{r}_1(t) or s=3s = -3 into r2(s)\mathbf{r}_2(s) to find the coordinates of PP.

Using r1(t)\mathbf{r}_1(t): r1(2)=2,4+(2),7+3(2)=2,2,1\mathbf{r}_1(-2) = \langle -2, 4 + (-2), 7 + 3(-2) \rangle = \langle -2, 2, 1 \rangle

Thus, the point of intersection is:

P=2,2,1P = \langle -2, 2, 1 \rangle

Would you like further details on the steps? Here are some additional related questions:

  1. How do you find the parametric equations for a line given two points?
  2. Can lines that don't intersect still have the same direction vectors?
  3. What if two lines are parallel but do not intersect? How would you handle that case?
  4. How can we check if two lines are skew (not intersecting and not parallel)?
  5. How would the solution change if the lines were defined in 2D instead of 3D?

Tip: When solving systems of equations with parametric variables, always check for consistency across all components to ensure there is an intersection point.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Parametric Equations
System of Linear Equations

Formulas

Parametric equation of a line: r(t) = r0 + t*v
Solving systems of equations to find intersection points

Theorems

None

Suitable Grade Level

Grades 10-12