 The distance between two points can be found using the distance formula. We could find the distance between a point and a line or a curve using a formula, but we won't. Remember, learn concepts, not formulas. Of course, a problem exists whether or not we know how to solve it, so let's try to find the distance between a point and a line given in vector form. A useful idea in math and in life, start with what you know. So, let's start with what you know. We know how to find the distance between two points, and so the distance between the point 3, 1, 7, and some point x, y, z on the line is... Now as the point of the line changes, the distance changes. And what we're calling the distance between the point and the line is the minimum value of the distance. So finding the distance corresponds to solving an optimization problem. So all we need to do is to solve an optimization problem in three variables, which we don't know how to do. While we don't know how to solve optimization problems in three variables, we do know that we have the vector equation for our line, and the vector equation corresponds to the parametric equation for x, y, and z. And this allows us to rewrite our distance formula as a function of t only. So our distance was substituting and simplifying. And so we get the distance between the point and the line given by. Now this is an optimization problem, so we'll differentiate. And while we could find the derivative directly, we'll square it to eliminate the square root and use implicit differentiation. And so we find the critical values will be where our denominator is zero or our numerator is zero. Since the point isn't on the line, s, which is the distance between the point and the line, can't be zero. And so the critical values will be the solutions to numerator equals zero. Solving gives us, and that gives us t equals 10 thirds. Since we want the distance, we can substitute into the equation for distance and find, and this will be the least distance, which will be the distance between the point and the line. The advantage of this approach is that it can be applied to any space curve. For example, let's consider this curve given parametrically, and let's try to find the point on the curve closest to the origin. So the distance between the origin and a point on the curve is given by. Again, we want to find the critical values, so we'll go ahead and work with s squared and implicit differentiation, and we find. And again, s can't be zero, so the only critical values are the solutions to this equation. Since this is a fifth degree equation, we need to solve it numerically. So we can use a tool like Wolfram Alpha, and it turns out the only real solution is. And we note that our coordinates will go to infinity as t goes to infinity, and because of that, the distance has no maximum value, so our critical value has to correspond to a least distance. So we can substitute this t value into our formula and find the least distance, and we can also substitute this into our parametric equations to find the actual coordinate of the point that is closest to the origin. Thank you.