# Tangent Line MultiVar Calc Problem

## Homework Statement

Find the slope of the tangent line to the curve of intersection of the vertical plane x-y+1= 0 and the surface z=x^2+ y^2 at the point (1, 2, 5) .

## The Attempt at a Solution

Normal Vector 1 is i-j
Normal Vector 2 is 2i+4j-k
Cross them to get i+j+6k
then turn that vector into a unit vector and after that I am unsure what to do

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lanedance
Homework Helper
first you need to decide what exactly you mean by slope...

Something like dz/du, where du is position change projected on the xy-plane, and dz is the change in vertical position. This lines up with the usual notion of slope.

If you are confident in you tangent vector (which i haven't checked) call it t:
t = xi+yj+zk

then the slope as defined above will be given by the ratio of z component to the projection on the xy plane
slope = z/(sqrt(x^2 + y^2))

Another easy way to find the tangent vector direction would be to parameterise the line in terms of t=x, giving r(t). Then differentiate in terms of t.