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Problem 589

# Give an example to show that two non-parallel lines in $$\mathrm{R}^{3}$$ need not intersect.

Expert verified
An example of two non-parallel lines in $$\mathrm{R}^{3}$$ that do not intersect are $$L_{1}: (x, y, z) = (1, 2, 3) + t(1, -1, 0)$$ and $$L_{2}: (x, y, z) = (2, 4, 1) + s(1, 1, -1)$$. The direction vectors for these lines are $$\vec{a} = (1, -1, 0)$$ and $$\vec{b} = (1, 1, -1)$$, which are not proportional, confirming that the lines are non-parallel. Solving for parameters t and s to check for intersection points, we find that they do not have a unique intersection point, indicating that the lines do not intersect. This example shows that in $$\mathrm{R}^{3}$$, two non-parallel lines do not necessarily intersect.
See the step by step solution

## Step 1: Choosing two non-parallel lines in $$\mathrm{R}^{3}$$

To choose two non-parallel lines, we need to make sure that their direction vectors are not proportional, meaning that there isn't a single constant that can be multiplied by one direction vector to obtain the other. Let's choose two lines in $$\mathrm{R}^{3}$$ with the following parametric equations: $$L_{1}: (x, y, z) = (1, 2, 3) + t(1, -1, 0)$$ $$L_{2}: (x, y, z) = (2, 4, 1) + s(1, 1, -1)$$

## Step 2: Check if the lines are non-parallel

To check if the lines are non-parallel, we will see if their direction vectors are not proportional. The direction vectors for the lines are: $$\vec{a} = (1, -1, 0)$$ for line $$L_{1}$$ and $$\vec{b} = (1, 1, -1)$$ for line $$L_{2}$$. No constant can be multiplied by the individual components of either vector to obtain the other. Therefore, since these direction vectors are not proportional, the lines are non-parallel.

## Step 3: Check if the lines intersect

To check if the lines intersect, we need to see if there exists a point that lies on both lines. Set the equations of the two lines equal to each other and try to solve for parameters t and s: $$(1, 2, 3) + t(1, -1, 0) = (2, 4, 1) + s(1, 1, -1)$$ This gives us the following system of equations: 1. $$1 + t = 2 + s$$ 2. $$2 - t = 4 + s$$ 3. $$3 = 1 - s$$ From equation 3, we can see that $$s = -2$$. Plugging this value into equations 1 and 2: 1. $$1 + t = 2 - 2 \Rightarrow t = -1$$ 2. $$2 - t = 4 - 2 \Rightarrow t = 0$$ As we got two different values of t corresponding to the same value of s (i.e., the t value was not unique), we can conclude there is no intersection point, and hence, no point where the two lines overlap. We have shown that the two lines $$L_{1}$$ and $$L_{2}$$ are non-parallel, and they do not intersect. This example demonstrates that two non-parallel lines in $$\mathrm{R}^{3}$$ do not necessarily intersect.

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