# Vector perpendicular

• Apr 20th 2011, 02:16 AM
johnsy123
Vector perpendicular
Find the vector perpendicular to the line 3x-5y+2=0.
• Apr 20th 2011, 02:22 AM
Plato
Quote:

Originally Posted by johnsy123
Find the vector perpendicular to the line 3x-5y+2=0.

The line 5x+3y+2=0 is perpendicular to that line.
• Apr 20th 2011, 05:31 AM
Prove It
Quote:

Originally Posted by johnsy123
Find the vector perpendicular to the line 3x-5y+2=0.

Write in the form y = mx + c, i.e. y = (3/5)x + 2/5, which has gradient = 3/5.

The gradients of perpendicular lines have a product of -1. So the gradient of your perpendicular is -5/3.

So any line in the family of lines y = (-5/3)x + c will be perpendicular. For simplicity we'll just consider y = (-5/3)x.

If you note from the slope that this perpendicular line moves down 5 units as you move to the right 3 units, then you can write this line in its vector form t(3i - 5j), where t is an arbitrary constant. If you know a point that this line goes through, say (x_0, y_0), then the line can be written as (x_0 + 3t)i + (y_0 - 5t)j.
• Apr 20th 2011, 06:16 AM
earboth
Quote:

Originally Posted by johnsy123
Find the vector perpendicular to the line 3x-5y+2=0.

If the vector (x, y) denotes the staionary vector of any point on the given line then the equation of the line is the result of a scalar product:

(3, -5) . (x, y) = -2

where the vector (3, -5) is the normal vector, that means it is perpendicular to the line.