# Probably quite simple but i'm stuck

• .....
In summary, the conversation is about finding the minimum distance between a point and a line in R^3. The approach suggested involves finding a point on the line and a vector between that point and the given point, then using the scalar projection to find the distance. The main question is how to find the normal vector of the line, and different approaches are discussed. One approach involves finding two points in cartesian coordinates and using the distance formula along with calculus to minimize the distance function. Another approach involves finding the point on the line where the normal vector passing through the given point cuts it. In the end, it is suggested to use the fact that PQ, the vector between the two points, is normal to the given line to solve the problem.
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I'm stuck on this question... if someone could give me a hand that'd be great..

What is the minimum distance between the point (3,-2,4) and the line defined by

x = 1 + t
y = 4 - 3t
z = -2 + 2t

My approach was to find a point on the line by letting t = 1

I got (2, 1, 0)

then I found a vector between this point and the one given:

v = i -3j + 4k

Here's a picture to clarify:

Then the distance would be the scalar projection of this vector onto the normal vecor of the line

But I'm not sure how to find the normal to the line.

Could someone show me how?

And if there's a simpler approach could you show me that too?

Thanks.

Geometrically,can u see what is the minimum distance...?Come on,it's something i learned when i was 12...

Daniel.

A single line in R^3 doesn't have a single normal vector, since it lies on an infinite number of planes.

The way I would try this to find two points in cartesian coordinates and find the line equation in those coordinates. Then using the distance formula and some max/min calculus, minimize the distance function.

here is what i would try

$$L = \sqrt{(x_0-x_1)^2 +(y_0-y_1)^2 (z_0-z_1)^2$$
$$L = \sqrt{(1+t-3)^2 +(4-3t+2)^2 (-2+2t-4)^2}$$
$$= \sqrt{(t-2)^2 +(-3t+6)^2 (2t-6)^2}$$
$$= |t-2|\sqrt{36t^2-216t+325}$$

Then take the derivative in respect to t. Find the critical points, and then pick the minimum one.

why not do something simpler? all you need to do is to find the point Q on the line where the normal vector passing through (3,-2,4) cuts it...that point looks like (1+t,4-3t,-2+2t) for some t. that is the t you need to find...and to do that use the fact that PQ is normal to the given line...

^^that works, i got it.

thanks a lot mansi.

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