Question

Find the distance between the point and the line given by the set of parametric equations. (-2,1,3)\(; \quad x=1-t, \quad y=2+t, \quad z=-2 t\)

Short answer

The minimum distance from the given point to the line can be obtained by applying the distance formula and using calculus to minimize the resulting expression.

Step-by-step solution

Find a point on the line

The line is defined by parametric equations, meaning each coordinate (x, y, z) is expressed as a function of t. Therefore, any point on the line can be found by finding (x, y, z) for an arbitrary value of t.

Derive the distance formula

The distance between two points in 3D space is given by the equation \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\). Where \(x_1, y_1, z_1\) are the coordinates of the given point and \(x_2, y_2, z_2\) are the coordinates of the point on the line in terms of t.

Minimize the distance

In order to find the minimum distance, differentiate the distance with respect to t, then set the derivative equal to zero and solve for t. This will yield the value of t where the distance from the line to the point is minimized.

Compute the minimum distance

Substituting the value of t that minimizes the distance into the equation obtained in Step 2 yields the minimum distance from the point to the line.

Key concepts

Parametric Equations

Parametric equations are a powerful way to describe a line in three-dimensional space. They're like a set of instructions that tell you how to find any point on a line by giving a single input, commonly known as a parameter. In our exercise, this parameter is represented by the letter 't'.

Think of 't' as a time marker on a journey along the line; as 't' changes, you move to different points. For the line in our exercise, the parametric equations are:

• \( x = 1 - t \) for the x-coordinate,
• \( y = 2 + t \) for the y-coordinate, and
• \( z = -2t \) for the z-coordinate.

With these equations, you can visualize the movement along the line in 3D by plugging in various values of 't'. This is especially useful when you can't easily sketch the line on graph paper because it's in three dimensions, not just two.

3D Distance Formula

If you've ever wondered how far apart two points are in space, the 3D distance formula is your go-to tool. Similar to finding the distance between two points on a flat surface, the 3D distance formula considers depth as well.

For two points \( P_1(x_1, y_1, z_1) \) and \( P_2(x_2, y_2, z_2) \) in space, the distance 'D' between them is calculated with the following equation: \( D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \). To find the distance between a point and a line, as in our exercise, you'd use the parametric equations of the line to express \( P_2 \) as \( (x, y, z) = (1 - t, 2 + t, -2t) \) and then plug these expressions, along with the coordinates of the given point \( P_1(-2, 1, 3) \) into the formula.

Differentiation

Differentiation is a mathematical operation that finds how a function changes as its input changes—essentially, it's the process of finding the rate of change. In the context of our exercise, after applying the 3D distance formula, we obtain a function that gives us the distance 'D' between the point and the line for any 't'.

But we want to find the point on the line that is closest to the given point, which means finding 't' where this distance is at its smallest. This is where differentiation comes into play. By differentiating the distance function with respect to 't' and setting the derivative equal to zero, we find the critical points. These critical points are the candidates for the minimum and maximum distances. Normally, we'd also use the second derivative test or other methods to confirm the minimum, but in this case, we are looking for the single minimum point, so the critical point where the first derivative is zero gives us the value of 't' for the closest approach.