# Thread: Vector in 3 space help

1. ## Vector in 3 space help

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

2. Originally Posted by Sherman60
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
i don't fully understand what you want. if you want a point not on the vector, why not just pick one. there are infinitely many. find a point on the vector, and just change one of the coordinates. you will get another vector in the same plane (the plane depends on which component you change)

3. 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.

4. Originally Posted by Sherman60
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.
ok, you should have just said the problem to begin with.

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 $ax + by + cz = d$. where $\vec n = \left< a,b,c \right>$ is the normal vector. the normal vector is perpendicular to the plane. no need for cross-products here

to find $d$. plug in $a,b,c,x,y,z$ and solve for it. you know $x,y,$ and $z$ from the point the line passes through

5. 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

6. Originally Posted by Sherman60
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
no, you take cross product if you are given two vectors in the plane. then you take the cross product to find the normal (which would be orthogonal to the two vectors).

a line can be written as $(x,y,z) = (x_0,y_0,z_0) + t \left< a,b,c \right>$

where $(x_0,y_0,z_0)$ is a point the line passes through and $\left< a,b,c \right>$ 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