How do you find a point on a line bisecting an angle in three-dimensional space?Given the x, y, z coordinates of three points P 1 , P 2 , P 3 with the angle between them being ∠ P 1 P 2 P 3 , how do you find a point, say at a distance of 1 from P 2 , on the line that bisects the angle?I know from the angle bisector theorem that the point must be equidistant from the the vector ( P 3 − P 2 ) and ( P 1 − P 2 ), but I can't seem to figure out how to find such a point in 3-space.

Question

How do you find a point on a line bisecting an angle in three-dimensional space?Given the x, y, z coordinates of three points       P    1    ,      P    2    ,      P    3   with the angle between them being   ∠      P    1        P    2        P    3  , how do you find a point, say at a distance of 1 from       P    2  , on the line that bisects the angle?I know from the angle bisector theorem that the point must be equidistant from the the vector   (      P    3    −      P    2    ) and   (      P    1    −      P    2    ), but I can&#039;t seem to figure out how to find such a point in 3-space.

Paola Mercer · Accepted Answer

Step 1Assume       P    1    ,      P    2    ,   and       P    3   aren&#039;t collinear. Put       v    1    =      P    1    −      P    2   and       v    2    =      P    3    −      P    2  . You need to find   v  ∈  span  {      v    1    ,      v    2    } satisfying the relationship                     v        1            ⋅      v                      |                    |                    v        1                    |                    |              =                    v        2            ⋅      v                      |                    |                    v        2                    |                    |            . If you write   v  =      c    1        v    1    +      c    2        v    2   you&#039;ll immediately recognize that                     v        1            ⋅      v                      |                    |                    v        1                    |                    |              =                    v        2            ⋅      v                      |                    |                    v        2                    |                    |                ⟺        c    2    =      c    1              (                          |                    |                    v        1                    |                    |            −                                    v            1                    ⋅                      v            2                                                |                                |                                v            2                                |                                |                                              |                    |                    v        2                    |                    |            −                                    v            1                    ⋅                      v            2                                                |                                |                                v            1                                |                                |                                          )      Step 2So if we assign       c    1    =  1 and       c    2    =                    |                    |                    v        1                    |                    |            −                                    v            1                    ⋅                      v            2                                                |                                |                                v            2                                |                                |                                              |                    |                    v        2                    |                    |            −                                    v            1                    ⋅                      v            2                                                |                                |                                v            1                                |                                |                               we see   v  =      v    1    +            (                          |                    |                    v        1                    |                    |            −                                    v            1                    ⋅                      v            2                                                |                                |                                v            2                                |                                |                                              |                    |                    v        2                    |                    |            −                                    v            1                    ⋅                      v            2                                                |                                |                                v            1                                |                                |                                          )            v    2   and the line   l  (  t  )  =      P    2    +  t  v bisects   ∠            P      1              P      2              P      3      . Notice how   l            (            1                  |                    |            v              |                    |                        )       is a point on this bisector that lands one unit away from       P    2  .

Paulkenyo · Accepted Answer

Step 1The sum of the two unit vectors       v    =                              P          2                          P          1                    →            /        |        P    2        P    1        |      +                                P          2                          P          3                    →            /        |        P    2        P    3        |   is a vector lying on the bisector of the angle between them.Step 2Make v unitary, multiply it by the distance d you want from       P    2   and add to                     O                  P          2                    →      .

Given the x, y, z coordinates of three points P_1, P_2, P_3 with the angle between them being angle P_1 P_2 P_3, how do you find a point, say at a distance of 1 from P_2, on the line that bisects the angle?

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