Question

Let L be the line with parametric equations

x = −5−3t
y = −4+2t
z = −2+2t
Find the shortest distance d from the point P0(1, −3, 5) to L, and the point Q on L that is closest to P0.

Answer 1

To find the shortest distance from a point to a line, we need to find the perpendicular distance from the point to the line. The point on the line that is closest to (1, -3, 5) is (-5, -4, -2).

Step-by-step explanation:

To find the shortest distance from a point to a line, we need to find the perpendicular distance from the point to the line. The direction vector of the line is given by <−3, 2, 2> and a point on the line is (-5, -4, -2). We can find the vector equation of the line as follows: <x, y, z> = <-5, -4, -2> + t<−3, 2, 2>.

Now we can find the equation of the plane perpendicular to the line passing through the point (1, -3, 5). The normal vector of the plane is the direction vector of the line, which is <−3, 2, 2>. So the equation of the plane is −3(x−1)+2(y+3)+2(z−5) = 0.

The shortest distance between the plane and the point is the perpendicular distance. Therefore, we substitute the coordinates of the point into the equation of the plane to find the shortest distance. After substituting the coordinates, we have: −3(1−1)+2(−3+3)+2(5−5)=0. Therefore, the shortest distance from the point (1, -3, 5) to the line is 0.

The point on the line that is closest to (1, -3, 5) can be found by solving the equation of the line and substituting the t value into the equation. After solving, we find that the point on the line is (-5, -4, -2). So the point Q on the line that is closest to (1, -3, 5) is (-5, -4, -2).