Question

I'm misunderstanding some elementary concept. Question 2 - http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-c-parametric-equations-for-curves/session-15-equations-of-lines/MIT18_02SC_pb_17_comb.pdf

I'm failing to follow the reasoning where it says the line of intersection is perpendicular to both normals.

Now, this seems to assume the coefficients of : (1,1,1) and (1,2,3) are both normals. However, I thought by the very definition of the dot product, two vectors are perpendicular (ie: one is normal) if and only if their

Answer 1

dot product is 0. Clearly in this question the two dot products are not equal to 0, they are equal to 1 and 2 respectively. How then can we treat these two coefficient vectors as being normals when their respective dot products with <x,y,z> does not equal 0?

I would like to both be shown where my reasoning is wrong and to also gain a full geometric visualization of what is happening here.

Answer 2

sorry I am not good with these

Answer 3

No problem, me neither unfortunately.

Answer 4

@dan815 can you take this?

Answer 5

you're confusing between "point" and "line"
any line on a plane is perpendicular to its normal vector

\(P_0\) is a point, not a line.

Answer 6

So yes, say we pick a point on the line P0, that is located on both planes, this must be part of their line of intersection. But what I'm referring to is the dot product of <x,y,z> . < 1,1,1> = 1 and <x,y,z> . <1,2,3> = 2

If <1,2,3> were normal to <x,y,z> shouldn't the dot product be 0?

Answer 7

you take dot product of normal with a line
not with a point

Answer 8

Here <x,y,z> refer to the position vector of any point on the plane dr

Answer 9

If you take a vector parallel to the plane that would be normal with the normal...

Answer 10

aren't you trying to do \(\large P_0 \bullet \text{n}\)

Answer 11

Ahh yes it seems so.

Answer 12

which makes no sense, i hope you see it by now :)

Answer 13

So the cross product then of two normals will give the perpendicular vector connecting them?

Answer 14

cross product of the normals gives the "direction vector" of the required line

Answer 15

just so you get some practice, try working another point on the intersection of given planes by setting \(z=1\)

call that point \(P_1\) and take the dot product of \(\overline{P_0P_1}\) with any of the normals to the planes. you should get 0.

Answer 16

Ahh yes, so say P1 = (1,-1,1). PoP1 = <1,-2,1> . <1,1,1> or <1,-2,1> . <1,2,3> = 0

Answer 17

Excellent!

Answer 18

Makes sense now, thanks for your help rational.

Answer 19

yw!