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

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

we'll equate the coordinates for each component \( x \), \( y \), and \( z \) in terms of the parameters \( t \) and \( u \), and then solve for \( t \) and \( u \) where the coordinates match.

### Step 1: Set up the equations for \( x \), \( y \), and \( z \)

From \(\vec{v}(t)\):
- \( x = 5t + 1 \)
- \( y = 9t + 1 \)
- \( z = -9t + 1 \)

From \(\vec{w}(u)\):
- \( x = -3u - 4 \)
- \( y = -4u - 8 \)
- \( z = 2u + 10 \)

### Step 2: Set up the system of equations

For intersection, we equate the coordinates:

1. \( 5t + 1 = -3u - 4 \)
2. \( 9t + 1 = -4u - 8 \)
3. \( -9t + 1 = 2u + 10 \)

### Step 3: Solve the system

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

#### 1. Solve for \( u \) and \( t \) from equations (1) and (2):

From equation (1):
\[
5t + 1 = -3u - 4
\Rightarrow 5t + 3u = -5
\]

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?

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### 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.

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

This guide explains how to find the intersection point of two lines in 3D space using vector algebra and parametric equations. The solution involves setting up and solving a system of equations for the x, y, and z components. Perfect for high school students learning advanced algebra and vector calculus.

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