Find Parametric Equation for Line Parallel to Two Planes

[knowLittle]

Homework Statement

Find parametric equations for the line which passes through the point (1; 2; 3)
and is parallel to both of the planes 3x + y + 5z = 4 and z = 1 -2x.

I have seen the result for this problem, but it's different than mine. I'm not sure, what I'm doing wrong. Please, help.

The Attempt at a Solution

If the line is parallel to two planes, then the line must be orthogonal to both planes' normals.
First, I notice that the planes' normal are:
n1=(3,1,5) and n2=(2,0,1)

Then, the direction vector of the line is p=<a,b,c>
and since, the vector and the normals are orthogonal.
*Dot Products*
p.n1=0
p.n2=0

I get :
3a+b+5c=0
2a+0+c=0
Using determinants, I find that the direction of the line is: 1i+7j-2k
{x-x_{1}} over {1} ={y-y_{1}} over {7} ={z-z_{1}} over {-2} =t
or
(x-1)=(y+2)/7 =(z-3)/-2 =t ...where the values of the point passed are plugged in on the numerator.
So, my parametric equation yields:
x=t+1
y=7t-2
z=-2t+3

 

[tiny-tim]

welcome to pf!

hi knowLittle! welcome to pf!

Find parametric equations for the line which passes through the point (1; 2; 3)
and is parallel to both of the planes 3x + y + 5z = 4 and z = 1 -2x.
…
So, my parametric equation yields:
x=t+1
y=7t-2
z=-2t+3

looks ok (apart from the -2 in the y) …

maybe the book is using a different parameter (eg t+1)

what answer does the book give?

 

[LCKurtz]

And you might notice that a quick way to get a direction vector along the line of intersection is to take the cross product of the normals.

 

[knowLittle]

This is the book's answer. It's very short and unclear.

1a: Take cross products of the displacement vectors to get a normal to plane P:
Cross [{0, 1, 2} - {1, 0, -1} , {1, 2, 3} - {0, 1, 2}]
{-2, 4, -2}
That gives an equation of the plane from the normal vector and a point:
-2x + 4(y - 1) - 2 (z - 2) ==0
-2 x + 4 (-1 + y) - 2 (-2 + z)==0

Also, I don't see any problem in the substraction of "2". It's the result of the determinant.

 

[tiny-tim]

I'm confused … that seems to be a different question

 

[knowLittle]

And you might notice that a quick way to get a direction vector along the line of intersection is to take the cross product of the normals.

I don't understand the cross product shown on the book.
My cross product of the normals give: 1i+ 7j-2k , where normal1=<3,1,5> && normal2=,2,0,1>

 

[knowLittle]

I think that you are right. The pages, where uploaded incorrectly by the staff. Wow.
Thank you all.

 

[knowLittle]

I think that you are right. The pages were uploaded incorrectly by the staff. Wow.
Thank you all.