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

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- Apr 20th 2011, 02:16 AMjohnsy123Vector perpendicular
Find the vector perpendicular to the line 3x-5y+2=0.

- Apr 20th 2011, 02:22 AMPlato
- Apr 20th 2011, 05:31 AMProve It
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(3**i**- 5**j**), 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 AMearboth