Math Problem Statement

Find the point of intersection for the lines \( \vec{v}(t) = (5, 9, -9)t + (1, 1, 1) \) and \( \vec{w}(u) = (-3, -4, 2)u + (-4, -8, 10) \).

Solution

To find the point of intersection of the two lines given by:

v⃗(t)=(5,9,−9)t+(1,1,1)\vec{v}(t) = (5, 9, -9)t + (1, 1, 1) and w⃗(u)=(−3,−4,2)u+(−4,−8,10),\vec{w}(u) = (-3, -4, 2)u + (-4, -8, 10),

we'll equate the coordinates for each component xx, yy, and zz in terms of the parameters tt and uu, and then solve for tt and uu where the coordinates match.

Step 1: Set up the equations for xx, yy, and zz

From v⃗(t)\vec{v}(t):

  • x=5t+1x = 5t + 1
  • y=9t+1y = 9t + 1
  • z=−9t+1z = -9t + 1

From w⃗(u)\vec{w}(u):

  • x=−3u−4x = -3u - 4
  • y=−4u−8y = -4u - 8
  • z=2u+10z = 2u + 10

Step 2: Set up the system of equations

For intersection, we equate the coordinates:

  1. 5t+1=−3u−45t + 1 = -3u - 4
  2. 9t+1=−4u−89t + 1 = -4u - 8
  3. −9t+1=2u+10-9t + 1 = 2u + 10

Step 3: Solve the system

Let's solve this system step-by-step.

1. Solve for uu and tt from equations (1) and (2):

From equation (1):

\Rightarrow 5t + 3u = -5$$ From equation (2): $$9t + 1 = -4u - 8 \Rightarrow 9t + 4u = -9$$ Now we have: 1. $$5t + 3u = -5$$ 2. $$9t + 4u = -9$$ Multiply the first equation by 4 and the second by 3 to eliminate $$u$$: - $$20t + 12u = -20$$ - $$27t + 12u = -27$$ Subtract the first from the second: $$7t = -7 \Rightarrow t = -1$$ #### 2. Substitute $$t = -1$$ into one of the original equations to find $$u$$: Using $$5t + 3u = -5$$: $$5(-1) + 3u = -5 \Rightarrow -5 + 3u = -5 \Rightarrow 3u = 0 \Rightarrow u = 0$$ ### Step 4: Find the intersection point by substituting $$t = -1$$ into $$\vec{v}(t)$$ or $$u = 0$$ into $$\vec{w}(u)$$: Substitute $$t = -1$$ into $$\vec{v}(t)$$: $$x = 5(-1) + 1 = -5 + 1 = -4$$ $$y = 9(-1) + 1 = -9 + 1 = -8$$ $$z = -9(-1) + 1 = 9 + 1 = 10$$ Thus, the point of intersection is: $$(x, y, z) = (-4, -8, 10)$$ ### Final Answer: - $$x = -4$$ - $$y = -8$$ - $$z = 10$$ Would you like further details or have any questions? --- ### Related Questions 1. How would the solution change if one of the direction vectors was parallel to the other? 2. What method could be used if there was no point of intersection (if the lines were skew)? 3. How can you determine if two lines in 3D space are parallel? 4. How can you find the angle between two lines in 3D space? 5. How would you find the intersection if the equations were given in parametric form? ### Tip In 3D, two lines may intersect, be parallel, or be skew. Checking if they intersect requires solving a system of equations based on their parametric forms.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Vector Algebra
Parametric Equations
Systems of Equations

Formulas

Parametric form of a line in 3D: \( \vec{r}(t) = \vec{d} t + \vec{p} \)
System of equations to equate vector components

Theorems

None specific, but relies on solving systems of linear equations for intersection

Suitable Grade Level

Grades 10-12

Related Recommendation

Finding Intersection of Two Lines in 3D Space Using Parametric Equations

Determine if Two Lines in 3D Space Intersect

Intersection of Line and Plane: Finding the Point of Intersection

Determine Intersection of Two Parametric Lines in 3D

Finding the Intersection Point of Two Parametric Lines in 3D