Vectors: show line x=0,y=t,z=2t par. to plane 2x-10y+5z-1=0

[Spoon-]

Hey, my question is:

If the positive z-axis points up, show that the line x = 0, y = t, z = 2t is parallel to and below the plane 2x - 10y + 5z - 1 =0.

I got the part finding that it's parallel by finding the dot product of the normal and direction vector. But I have no clue how to show that it's below the plane.

Thanks.

 

[pka]

Re: Vectors and Scalers of a plane

If the positive z-axis points up, show that the line x = 0, y = t, z = 2t is parallel to and below the plane 2x - 10y + 5z - 1 =0.

The point (0,0,0) is on the line, correct?
In some sense is that point "below" the plane?

 

[Deleted member 4993]

Hey, my question is:

If the positive z-axis points up, show that the line x = 0, y = t, z = 2t is parallel to and below the plane 2x - 10y + 5z - 1 =0.

I got the part finding that it's parallel by finding the dot product of the normal and direction vector. But I have no clue how to show that it's below the plane.

Thanks.

For points on the plane,

when x=0 & y=0 - what is the 'z' value of the point (on the plane) - a positive value - correct?

when x=0 & z=0 - what is the 'y' value of the point (on the plane) - a positive value - correct?

when y=0 & z=0 - what is the 'x' value of the point (on the plane) - a positive value - correct?

This shows intuitively that the plane is above the origin.

Now (depending on your class notes - methods taught) you may need to show this in through normal (directional) vectors.