Question

Find the distance between the point and the line given by the set of parametric equations. (1,-2,4)\(; \quad x=2 t, \quad y=t-3, \quad z=2 t+2\)

Short answer

The distance between the point (1,-2,4) and the line given by the parametric equations \(x=2t, y=t-3, z=2t+2\) is 1.

Step-by-step solution

Identify the given point

The given point in the exercise is (1,-2,4)

Define the parametric equations for the line

The line in 3D space is defined by the parametric equations \(x=2t\), \(y=t-3\), and \(z=2t+2\)

Find the point on the line closest to the given point by setting the equations equal

Set the coordinates of the given point equal to the expressions in the parametric equations to find the corresponding value of \(t\). So, from \(x=2t\) we have \(t=1/2\), from \(y=t-3\) we get \(t=-2+3=1\), and from \(z=2t+2\) we have \(t=(4-2)/2=1\).

Reconcile differing values of t

The solution for \(t\) is differing depending on the parametric equation used. Here, \(t=1/2\) from equation \(x=2t\), and \(t=1\) from equations \(y=t-3\) and \(z=2t+2\). This suggests that there is no point on the line that matches exactly with the given point. In such cases, we choose the consistent value among the different solutions, which is \(t=1\).

Substitute t back into the parametric equations

Substitute \(t=1\) into the parametric equations to find the point on the line that is closest to the given point. You get \(x=2(1)=2\), \(y=1-3=-2\), and \(z=2(1)+2=4\), so the point is (2,-2,4)

Use the distance formula

The distance between two points (x1, y1, z1) and (x2, y2, z2) is given by \(\sqrt{(x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2}\). Substituting the given point (1,-2,4) and the point found on the line (2,-2,4) into the formula gives \(\sqrt{(2-1)^2 + (-2+2)^2 + (4-4)^2} = \sqrt{1}\)