Consider the line segment in a (x,y) coordinate system connecting the points (2,3) and (5,4). Compute the Unit Vector normal to this line segment.
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Slope of the line through (2, 3) and (5, 4) = (4 - 3)/(5 - 2) = 1/3
Perpendicular lines have slopes that are negative reciprocals of each other so the perpendicular vector's slope = -1/(1/3) = -3
<1, -3> would be one vector with a slope of -3.
The unit vector for <1, -3> is
<1/(√(1^2 + (-3)^2), -3/(√(1^2) + (-3)^2)> =
<1/√10, -3/√10>
Perpendicular lines have slopes that are negative reciprocals of each other so the perpendicular vector's slope = -1/(1/3) = -3
<1, -3> would be one vector with a slope of -3.
The unit vector for <1, -3> is
<1/(√(1^2 + (-3)^2), -3/(√(1^2) + (-3)^2)> =
<1/√10, -3/√10>