Multivariable Calculus: Distance from a point to a plane

In this exercise, we find the distance from the point (1, 2, 3) to the plane 3x - y + 5z = 2 using geometric methods. We start by identifying another point on the plane and connect it to our given point with a vector v. We then project v onto a line perpendicular to the plane by projecting it on the normal vector n derived from the plane's equation. The distance is calculated as the absolute value of the dot product of v and n divided by the magnitude of N. This approach emphasizes understanding vector projections for solving various distance problems in geometry without relying on specific formulas.

#mathematics #math #vectorcalculus #multivariablecalculus #linesandplanes #iitjammathematics #calculus3

[thomashowe]

Honestly, it took me a couple of minutes to see that one does not need to find the point where “l” and the plane intersect. Thanks for the insight.

[mmd]

Hi, if we have more teachers like you, math would be accepted easier. My university math professor tells us "memory this and that", i cant im sorry. hahah

[sethsherman-ekpo]

Hey there. For some reason, I can no longer find your Cal 4 series of videos. Am I overlooking something, or are they no longer available? Thank you!

[mmd]

My approach was we have point (1, 2, 3) and we have normal vector (3, -1, 5) so i did this line (x, y, z)=(1, 2, 3)+k(3, -1, 5) intersected with the plane then calculated the distance between points.

[jackflash]

It isn't clear to me how you made the assumption that vector n had coordinates 3, -1, 5 . If you are using some 3D coordinate reference system, the coordinates 3, -1, 5 is just a point on the plane, not a vector. To determine vector n, surely one would need the coordinates of the tip end of vector n as well as the coordinates of its tail end (which is 3, -1, 5). Further, sq root 35 x sq root 35 = 35

[sethsherman-ekpo]

Sorry for bothering you, then. I thought I remembered seeing it before I left Africa. Probably just wishful thinking.