OK so I am in calc 3, and I need to know how to get the equation of two vectors in the same plane from just this:
L1 = (x-1)/7 = (y-2)/3 = z
I know how to get the vector and a point, but I need another point that is not on this vector.
Any help or techniques? Thanks
Thanks for the response, as you can see I am no english major, so it's probably that I'm not explaining it that well.
Here is the whole problem:
Find the equation of the plane perpendicular to Line 1 (L1 = (x-1)/7 = (y-2)/3 = z) and passing through the point (0,4,1)
Now I figured I'd have to find a point no on the given vector and take the cross product, thus finding the normal vector. I would then get the equation of the plane containing both the normal and the given point.
i suppose you can find the direction vector of the line. this will actually be the normal vector to your plane.
your plane is of the form . where is the normal vector. the normal vector is perpendicular to the plane. no need for cross-products here
to find . plug in and solve for it. you know and from the point the line passes through
Oh so this problem is rather simple, and I must be thinking too much about it.
I'm not sure what you mean by direction vector, but I was thinking of picking an arbitrary point, getting a vector, and then taking the cross product, thus finding the normal.
The rest I understand, though. Thanks again
a line can be written as
where is a point the line passes through and is the direction vector of the line. it describes the direction the line goes in according to the vector
here we were given a line. the normal vector to the plane perpendicular to it is the same as its direction vector
WARNING: by taking the cross product here, you are finding a vector in the plane, not perpendicular to it