To find the midpoint of a line AB,with the given points A and B

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To find the midpoint of line segment AB with points A=(4,3,2) and B=(-2,0,5), the correct midpoint is calculated as (1, 1.5, 3.5). The direction vector from the origin to this midpoint is derived as (1, 1.5, 3.5). To find the unit vector in this direction, it is expressed as (1i + 1.5j + 3.5k) divided by its magnitude. The magnitude is calculated as √(1^2 + 1.5^2 + 3.5^2). The final unit vector points from the origin towards the midpoint of line AB.
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Homework Statement


*)Given two points A=(4,3,2) and B=(-2,0,5).Find the unit vector extending from origin toward the midpoint of line AB


Homework Equations





The Attempt at a Solution



Direction vector to AB=(-6,-3,3) and hence direction vector to its midpoint is (-3,-1.5,1.5)
So unit vector in the direction of (-3,-1.5,1.5) is (-3i-1.5j+1.5k)/√(9+2.25+2.25).

Is this the way to do it??
 
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I doubt that you have begun this correctly, but my maths is very rusty.

Perhaps you should take another look at it by first calculating the midpoint of line AB.
 
Abel I Daniel said:

Homework Statement


*)Given two points A=(4,3,2) and B=(-2,0,5).Find the unit vector extending from origin toward the midpoint of line AB


Homework Equations





The Attempt at a Solution



Direction vector to AB=(-6,-3,3) and hence direction vector to its midpoint is (-3,-1.5,1.5)
So unit vector in the direction of (-3,-1.5,1.5) is (-3i-1.5j+1.5k)/√(9+2.25+2.25).

Is this the way to do it??

No. Look up the formula for the mid-point between two points.
 
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oh,i got wrong with the midpoint formula..so now the midpoint is (1,1.5,3.5).is it correct??
 
Abel I Daniel said:
oh,i got wrong with the midpoint formula..so now the midpoint is (1,1.5,3.5).is it correct??
That is the midpoint, yes.
 
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