Question

Find the distance between the point and the line given by the set of parametric equations. (Round your answer to three decimal places.)

(9, -7, 1); x = 2t, y = t − 3, z = 2t + 2.

Answer 1

The student needs to calculate the distance between a point (9, -7, 1) and a line with parametric equations x = 2t, y = t - 3, z = 2t + 2. This involves identifying a point on the line, finding the direction vector and the position vector, computing the cross product of those vectors, and applying the formula to find the distance.

Step-by-step explanation:

The student is interested in finding the distance between a point and a line described by parametric equations. The point in question is (9, -7, 1), and the line's parametric equations are x = 2t, y = t - 3, z = 2t + 2.

To find this distance, we can use the following steps:

1. First, identify a point on the line by choosing a value for t. As the value of t is not specified, we can choose t = 0 for simplicity, which gives us the point (0, -3, 2) on the line.
2. Next, find the direction vector of the line. This is given by the coefficients of t in the line's parametric equations, resulting in the vector = <2, 1, 2>.
3. Then, find the vector connecting the point on the line to the given point (position vector). This vector is (9 - 0, -7 - (-3), 1 - 2) = <9, -4, -1>.
4. Compute the cross product of and to find a vector that is perpendicular to both the direction vector of the line and the position vector from the line to the point.
5. The magnitude of this cross product gives us the area of a parallelogram with sides and . To find the distance, we divide this area by the magnitude of the direction vector .

The formula to compute the distance is thus: \[Distance = \frac \times \\\]